
Chapter 46 – The Nucleus
Page No 440:
Question 1:
If neutrons exert only attractive force, why don’t we have a nucleus containing neutrons alone?
Answer:
Nuclear forces are short range strong attractive forces that act between two protonproton, neutronproton and neutronneutron pairs. Two protons have strong nuclear forces between them and also exert electrostatic repulsion on each other. However, electrostatic forces are long ranged and have very less effect as compared to the strong nuclear forces.
So, in a nucleus (that is very small in dimension), there’s no such significance of repulsive force as compared to the strong attractive nuclear force. On the other hand, an atom contains electrons revolving around its nucleus. These electrons are kept in their orbit by the strong electrostatic force that is exerted on them by the protons present inside the nucleus. Hence, a nucleus contains protons as well as neutrons.
Question 2:
Consider two pairs of neutrons. In each pair, the separation between the neutrons is the same. Can the force between the neutrons have different magnitudes for the two pairs?
Answer:
Neutrons are chargeless particles and they exert only short range nuclear forces on each other. If we have two pairs of neutrons and the separation between them is same in both the pairs. The force between the neutrons will be of same magnitude for the two pairs until there is some other influence on any of them.
Question 3:
A molecule of hydrogen contains two protons and two electrons. The nuclear force between these two protons is always neglected while discussing the behaviour of a hydrogen molecule. Why?
Answer:
Inside the nucleus, two protons exert nuclear force on each other. These forces are shortranged (a few fm), strong and attractive forces. They also exert electrostatic repulsive force (longranged). While discussing the behaviour of a hydrogen molecule, the nuclear force between the two protons is always neglected. This is because the separation between the two protons in the molecule is
~70 pm which is much greater than the range of the nuclear force.
Question 4:
Is it easier to take out a nucleon (a) from carbon or from iron (b) from iron or from lead?
Answer:
Binding energy per nucleon of a nucleus is defined as the energy required to breakoff a nucleon from it.
(a) As the binding energy per nucleon of iron is more than that of carbon, it is easier to take out a nucleon from carbon than iron.
(b) As the binding energy per nucleon of iron is more than that of lead. Therefore, it is easier to take out a nucleon from lead as compared to iron.
Question 5:
Suppose we have 12 protons and 12 neutrons. We can assemble them to form either a ^{24}Mg nucleus or two ^{12}C nuclei. In which of the two cases more energy will be liberated?
Answer:
If we assemble 6 protons and 6 neutrons to form ^{12}C nucleus, 92.15 MeV (product of mass number and binding energy per nucleon of carbon12) of energy is released. Therefore, the energy released in the formation of two carbon nuclei is 184.3 MeV. On the other hand, when 12 protons and 12 neutrons are combined to form a ^{24}Mg atom, 198.25 MeV of energy (binding energy) is released. Hence, in case of â€‹^{24}Mg nucleus, more energy is liberated.
Question 6:
What is the difference between cathode rays and beta rays? When the two are travelling in space, can you make out which is the cathode ray and which is the beta ray?
Answer:
Cathode rays consist of electrons that are accelerated using electrodes. They do not carry high energy and do not harm human body. On the other hand, beta rays consist of highly energetic electrons that can even penetrate and damage human cells. Beta rays are produced by the decay of radioactive nuclei. If the two are travelling in space, they can be distinguished by the phenomenon named production of Bremsstrahlung radiation, which is produced by the deceleration of a high energy particle when deflected by another charged particle, leading to the emission of blue light. Only beta rays are capable of producing it.
Question 7:
If the nucleons of a nucleus are separated from each other, the total mass is increased. Where does this mass come from?
Answer:
When the nucleons of a nucleus are separated, a certain amount of energy is to be given to the nucleus, which is known as the binding energy.
Binding energy = [(Number of nucleons)
×(Mass of a nucleon) – (Mass of the nucleus)]
When the nucleons of a nucleus are separated, the increase in the total mass comes from the binding energy, which is given to the nucleus to breakoff its constituent nucleons as energy is related to mass by the relation given below.
E = Δmc^{2}
Question 8:
In beta decay, an electron (or a positron) is emitted by a nucleus. Does the remaining atom get oppositely charged?
Answer:
In beta decay, a neutron from the nucleus is converted to a proton releasing an electron and an antineutrino or a proton is converted to a neutron releasing a positron and a neutrino.
i.e. β decay: n→p+e+ν¯β+ decay: p→n+e++νSince the number of valence electrons present in the parent atom do not change, the remaining atom does not get oppositely charged. Instead, due to a change in the atomic number, there’s a formation of a new element.
Question 9:
When a boron nucleus
(B510)is bombarded by a neutron, a αparticle is emitted. Which nucleus will be formed as a result?
Answer:
It is given that when a boron nucleus
(B510)is bombarded by a neutron, an αparticle is emitted.
Let X nucleus be formed as a result of the bombardment.
According to the charge and mass conservation,
B510+n01→X+H24eCharge: 5 0 3 2Mass: 10 1 7 4The mass number of X should be 7 and its atomic number should be 3.
∴X=L37i
Question 10:
Does a nucleus lose mass when it suffers gamma decay?
Answer:
Gamma rays consist of photons that are produced when a nucleus from its excited state comes to its ground state releasing energy. Since gamma rays are chargeless and massless particles, the nucleus does not suffer any loss in mass during the gamma decay.
Question 11:
In a typical fission reaction, the nucleus is split into two middleweight nuclei of unequal masses. Which of the two (heavier or lighter) has greater kinetic energy? Which one has greater liner momentum?
Answer:
Two photons having equal liner momentum have equal wavelengths as here for both the photons the direction and magnitude of linear momentum will be same. For the rest of the options, magnitude will be same but nothing can be said about the direction of the photons.
Hence the correct option is D.
Question 12:
If three helium nuclei combine to form a carbon nucleus, energy is liberated. Why can’t helium nuclei combine on their own and minimise the energy?
Answer:
When three helium nuclei combine to form a carbon nucleus, energy is liberated. This energy is greater than the that liberated when these nuclei combine on their own. Hence, formation of carbon nucleus leads to much more stability as compared to the combination of three helium nuclei.
Question 1:
The mass of a neutral carbon atom in ground state is
(a) exact 12 u
(b) less than 12 u
(c) more than 12 u
(d) depends on the form of carbon such as graphite of charcoal.
Answer:
(a) exact 12 u
In nuclear physics, a unit used for measurement of mass is unified atomic mass unit, which is denoted by u.
It is defined such that
1 u =
112×(Mass of neutral carbon atom in its ground state)
Mass of neutral carbon atom in its ground state = 12 × 1 u = 12 u
Thus, the mass of neutral carbon atom in its ground state is exactly 12 u.
Question 2:
The mass number of a nucleus is equal to
(a) the number of neutrons in the nucleus
(b) the number of protons in the nucleus
(c) the number of nucleons in the nucleus
(d) none of them.
Answer:
(c) the number of nucleons in the nucleus
Mass number of a nucleus is defined as the sum of the number of neutron and protons present in the nucleus, i.e. the number of nucleons in the nucleus.
Question 3:
As compared to ^{12}C atom, ^{14}C atom has
(a) two extra protons and two extra electrons
(b) two extra protons but no extra electrons
(c) two extra neutrons and no extra electron
(d) two extra neutrons and two extra electron.
Answer:
(c) two extra neutrons and no extra electron
^{12}C and ^{14}C are the two isotopes of carbon atom that have same atomic number, but different mass numbers. This means that they have same number of protons and electrons, but different number of neutrons. Therefore, â€‹^{12â€‹}C has 6 protons, 6 electrons and 6 neutrons, whereas â€‹^{14}C has 6 electrons, 6 protons and 8 neutrons.
Question 4:
The mass number of a nucleus is
(a) always less than its atomic number
(b) always more than its atomic number
(c) equal to its atomic number
(d) sometimes more than and sometimes equal to its atomic number.
Answer:
(d) sometimes more than and sometimes equal to its atomic number
Mass number of a nucleus is defined as the sum of the number of neutron and protons present in the nucleus, i.e. the number of nucleons in the nucleus, whereas atomic number is equal to the number of protons present. Therefore, the atomic number is smaller than the mass number. But in the nucleus (like that of hydrogen ^{1}H_{â€‹1}), only protons are present. Due to this, the mass number is equal to the atomic number.
Question 5:
The graph of ln(R/R_{0}) versus ln A(R = radius of a nucleus and A = its mass number) is
(a) a straight line
(b) a parabola
(c) an ellipse
(d) none of them.
Answer:
(a) a straight line
The average nuclear radius (R) and the mass number of the element (A) has the following relation:
R=RoA13RRo=A13lnRRo=13ln ATherefore, the graph of ln(R/R0) versus ln A is a straight line passing through the origin with slope 1/3.
Question 6:
Let F_{pp}, F_{pn} and F_{nn} denote the magnitudes of the net force by a proton on a proton, by a proton on a neutron and by a neutron on a neutron respectively. neglect gravitational force. When the separation is 1 fm.
(a) F_{pp} > F_{pn} = F_{nn}
(b) F_{pp} = F_{pn} = F_{nn}
(c) F_{pp} > F_{pn} > F_{nn}
(d) F_{pp} < F_{pn} = F_{nn}
Answer:
(d) Fpp < Fpn = Fnn
Protons and neutrons are present inside the nucleus and they exert strong attractive nuclear forces on each other, which are equal in magnitude. Due to their positive charge, protons repel each other. Hence the net attractive force between two protons gets reduced, but the nuclear force is stronger than the electrostatic force at a separation of 1 fm.
∴ Fpp < Fpn = Fnn
Here, Fpp, Fpn and Fnn denote the magnitudes of the net force by a proton on a proton, by a proton on a neutron and by a neutron on a neutron, respectively.
Question 7:
Let F_{pp}, F_{pn} and F_{nn} denote the magnitudes of the nuclear force by a proton on a proton, by a proton on a neutron and by a neutron on a neutron respectively. When the separation is 1 fm,
(a) F_{pp} > F_{pn} = F_{nn}
(b) F_{pp} = F_{pn} = F_{nn}
(c) F_{pp} > F_{pn} > F_{nn}
(d) F_{pp} < F_{pn} = F_{nn}
Answer:
(b) F_{pp} = F_{pn} = F_{nn}
Protons and neutrons are present inside the nucleus and they exert strong attractive nuclear force on each other. These forces are equal in magnitude, irrespective of the charge present on the nucleons.
∴ F_{pp} = F_{pn} = F_{nn}
Here, F_{pp}, F_{pn} and F_{nn} denote the magnitudes of the nuclear force by a proton on a proton, by a proton on a neutron and by a neutron on a neutron, respectively.
Question 8:
Two protons are kept at a separation of 10 nm. Let F_{n} and F_{e} be the nuclear force and the electromagnetic force between them.
(a) F_{e} = F_{n}
(b) F_{e} >> F_{n}
(c) F_{e} << F_{n}
(d) F_{e} and F_{n} differ only slightly.
Answer:
(b) F_{e} >> F_{n}
Two protons exert strong attractive nuclear force and repulsive electrostatic force on each other. Nuclear forces are short range forces existing in the range of a few fms. Therefore, at a separation of 10 nm, the electromagnetic force is greater than the nuclear force, i.e. F_{e} >> F_{n}.
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Question 9:
As the mass number A increases, the binding energy per nucleon in a nucleus
(a) increases
(b) decreases
(c) remains the same
(d) varies in a way that depends on the actual value of A.
Answer:
(d) varies in a way that depends on the actual value of A
Binding energy per nucleon in a nucleus first increases with increasing mass number (A) and reaches a maximum of 8.7 MeV for A (50−80). Then, again it slowly starts decreasing with the increase in A and drops to the value of 7.5 MeV.
Question 10:
Which of the following is a wrong description of binding energy of a nucleus?
(a) It is the energy required to break a nucleus into its constituent nucleons.
(b) It is the energy made available when free nucleons combine to form a nucleus.
(c) It is the sum of the rest mass energies of its nucleons minus the rest mass energy of the nucleus.
(d) It is the sum of the kinetic energy of all the nucleons in the nucleus.
Answer:
(d) It is the sum of the kinetic energies of all the nucleons present in the nucleus.
Binding energy of a nucleus is defined as the energy required to break the nucleus into its constituents. It is also measured as the Qvalue of the breaking of nucleus, i.e. the difference between the rest energies of reactants (nucleus) and the products (nucleons) or the difference between the kinetic energies of the products and the reactants.
Question 11:
In one averagelife,
(a) half the active nuclei decay
(b) less than half the active nuclei decay
(c) more than half the active nuclei decay
(d) all the nuclei decay.
Answer:
(c) more than half the active nuclei decay
The average life is the mean life time for a nuclei to decay.
It is given as
τ=1λ=Τ120.693Here,
τ, λ and Τ12 are the average life, decay constant and half lifetime of the active nuclei, respectively. The value of the average lifetime comes to be more than the average lifetime. Therefore, in one average life, more than half the active nuclei decay.
Question 12:
In a radioactive decay, neither the atomic number nor the mass number changes. Which of the following particles is emitted in the decay?
(a) Proton
(b) Neutron
(c) Electron
(d) Photon
Answer:
(d) Photon
The atomic number and mass number of a nucleus is defined as the number of protons and the sum of the number of protons and neutrons present in the nucleus, respectively. Since in the decay, neither the atomic number nor the mass number change, it cannot be a betadecay (release of electron, proton or neutron). Hence, the particle emitted can only be a photon.
Question 13:
During a negative beta decay,
(a) an atomic electron is ejected
(b) an electron which is already present within the nucleus is ejected
(c) a neutron in the nucleus decays emitting an electron
(d) a proton in the nucleus decays emitting an electron.
Answer:
(c) a neutron in the nucleus decays emitting an electron
Negative beta decay is given as
n→p+e+ν¯
Neutron decays to produce proton, electron and antineutrino.
Question 14:
A freshly prepared radioactive source of halflife 2 h emits radiation of intensity which is 64 times the permissible safe level. The minimum time after which it would be possible to work safely with this source is
(a) 6 h
(b) 12 h
(c) 24 h
(d) 128 h.
Answer:
(b) 12 h
A freshly prepared radioactive source emits radiation of intensity that is 64 times the permissible level. This means that it is possible to work safely till 6 halflives (as 2^{6} = 64) of the radioactive source. Since the halflife of the source is 2h, the minimum time after which it would be possible to work safely with this source is 12 h.
Question 15:
The decay constant of a radioactive sample is λ. The halflife and the averagelife of the sample are respectively
(a) 1/λ and (In 2/λ)
(b) (In 2/λ) and 1/λ
(c) λ(In 2) and 1/λ
(d) λ/(In 2) and 1/λ.
Answer:
(b) (ln 2/λ) and 1/λ
The halflife of a radioactive sample (
t12) is defined as the time elapsed before half the active nuclei decays.
Let the initial number of the active nuclei present in the sample be N_{0}.
No2=Noeλt12⇒t12=ln 2λAverage life of the nuclei,
tav=SNo=1λHere, S is the sum of all the lives of all the N nuclei that were active at t = 0 and
λ is the decay constant of the sample.
Question 16:
An αparticle is bombarded on ^{14}N. As a result, a ^{17}O nucleus is formed and a particle is emitted. This particle is a
(a) neutron
(b) proton
(c) electron
(d) positron.
Answer:
(b) proton
If an alpha particle is bombarded on a nitrogen (N14) nucleus, an oxygen (O17) nucleus and a proton are released.
According to the conservation of mass and charge,
H24e+N714→O617+p11So, the emitted particle is a proton.
Question 17:
Ten grams of ^{57}Co kept in an open container betadecays with a halflife of 270 days. The weight of the material inside the container after 540 days will be very nearly
(a) 10 g
(b) 5 g
(c) 2.5 g
(d) 1.25 g.
Answer:
(a) 10 g
^{57}Co is undergoing beta decay, i.e. electron is being produced. But an electron has very less mass (9.11
×10^{31 }kg) as compared to the Co atom. Therefore, after 570 days, even though the atoms undergo large beta decay, the weight of the material in the container will be nearly 10 g.
Question 18:
Free ^{238}U nuclei kept in a train emit alpha particles. When the train is stationary and a uranium nucleus decays, a passenger measures that the separation between the alpha particle and the recoiling nucleus becomes x in time t after the decay. If a decay takes place when the train is moving at a uniform speed v, the distance between the alpha particle and the recoiling nucleus at a time t after the decay, as measured by the passenger will be
(a) x + vt
(b) x − vt
(c) x
(d) depends on the direction of the train.
Answer:
(c) x
When the train is stationary, the separation between the alpha particle and recoiling uranium nucleus is x in time t after the decay. Even if the decay is taking place in a moving train and the separation is measured by the passenger sitting in it, the separation between the alpha particle and nucleus will be x. This is because the observer is also moving with the same speed with which the alpha particle and recoiling nucleus are moving, i.e. they all are in the same frame that is moving at a uniform speed.
Question 19:
During a nuclear fission reaction,
(a) a heavy nucleus breaks into two fragments by itself a light nucleus bombarded by thermal neutrons breaks up
(b) a light nucleus bombarded by thermal neutrons breaks up
(c) a heavy nucleus bombarded by thermal neutrons breaks up
(d) two light nuclei combine to give a heavier nucleus and possible other products.
Answer:
(c) a heavy nucleus bombarded by thermal neutrons breaks up
In a nuclear reactor, a large fissile atomic nucleus like uranium235 absorbs a thermal neutron and undergoes a nuclear fission reaction. The heavy nucleus splits into two or more lighter nuclei releasing gamma radiation and free neutrons.
Question 1:
As the mass number A increases, which of the following quantities related to a nucleus do not change?
(a) Mass
(b) Volume
(c) Density
(d) Binding energy
Answer:
(c) Density
Radius of a nucleus with mass number A is given as
R=RoA13
Here, R_{o} = 1.2 fm
∴ Volume of the nucleus =
4πR33=4πRo3A3
This depends on A. With an increase in A, V increases proportionally.
Mass of the nucleus
≃ Am_{N}
Here, m_{N}_{ }is the mass of a nucleon.
Therefore, mass of the nucleus also increases with the increasing mass number.
Binding energy also depends on mass number (number of nucleons) as it is the difference between the total mass of the constituent nucleons and the nucleus. Therefore, it also varies with the changing mass number.
On the other hand,
Density =
MassVolume=AmN4πR33=AmN4πRo3A3=mN4πRo33=3mN4πRo3This is independent of A and hence does not change as mass number increases.
Question 2:
The heavier nuclei tend to have larger N/Z ratio because
(a) a neutron is heavier than a proton
(b) a neutron is an unstable particle
(c) a neutron does not exert electric repulsion
(d) Coulomb forces have longer range compared to the nuclear forces.
Answer:
(c) a neutron does not exert electric repulsion
(d) Coulomb forces have longer range compared to the nuclear forces
This is because in heavy nuclei, the N/Z ratio becomes larger in order to maintain their stability and reduce instability caused due to the repulsion among the protons.The neutrons exert only attractive shortrange nuclear forces on each other as well as on the neighbouring protons, whereas the protons exert attractive shortrange nuclear forces on each other as well as the electrostatic repulsive force. Thus, the nuclei with high mass number, in order to be stable, have large neutron to proton ratio (N/Z).
Question 3:
A free neutron decays to a proton but a free proton does not decay to a neutron. This is because
(a) neutron is a composite particle made of a proton and an electron whereas proton is a fundamental particle
(b) neutron is an uncharged particle whereas proton is a charged particle
(c) neutron has large rest mass than the proton
(d) weak forces can operate in a neutron but not in a proton.
Answer:
(c) neutron has large rest mass than the proton.
A nucleus is made up of two fundamental particles neutrons and protons. If a nucleus has more number of neutrons than what is needed to have stability, then neutrons decay into protons and if there’s an excess of protons, then they decay to form neutrons. Since a neutron has larger rest mass than a proton, the Qvalue of its decay reaction is positive and a free neutron decays to a proton, while an isolated proton cannot decay to a neutron as the Qvalue of its decay reaction is negative. Hence, it is physically not possible.
Question 4:
Consider a sample of a pure betaactive material.
(a) All the beta particles emitted have the same energy.
(b) The beta particles originally exist inside the nucleus and are ejected at the time of beta decay.
(c) The antineutrino emitted in a beta decay has zero mass and hence zero momentum.
(d) The active nucleus changes to one of its isobars after the beta decay.
Answer:
(d) The active nucleus changes to one of its isobars after the beta decay.
In a beta decay, either a neutron is converted to a proton or a proton is converted to a neutron such that the mass number does not change. Also,the number of the nucleons present in the nucleus remains the same. Thus, the active nucleus gets converted to one of its isobars after beta decay.
Question 5:
In which of the following decays the element does not change?
(a) αdecay
(b) β^{+}decay
(c) β^{−}decay
(d) γdecay
Answer:
(d) γdecay
â€‹In alpha particle decay, the unstable nucleus emits an alpha particle reducing its proton number Z by 4 and neutron number N by 2 such that the element gets changed.
XZA→YZ2A4+H24e
During β^{−}decayâ€‹, a neutron is converted to a protonâ€‹, an electron and an antineutrino, i.e. an active nucleus gets converted to one of its isobars and hence the element gets changed.
XZA→YZ+1A+e+ν¯
During β^{+}–decayâ€‹, a proton in the nucleus is converted to a neutronâ€‹, a positron and a neutrino in order to maintain the stability of the nucleus, i.e. an active nucleus gets converted to one of its isobars and hence the element gets changed.
XZA→YZ1A+β++ν
When a nucleus is in higher excited state or has excess of energy, it comes to the ground state in order to become stable and release energy in the form of electromagnetic radiation called gamma ray. Hence, the element in gamma decay doesn’t change.
Question 6:
In which of the following decays the atomic number decreases?
(a) αdecay
(b) β^{+}decay
(c) β^{−}decay
(d) γdecay
Answer:
(a) αdecay
(b) β^{+}decay
â€‹In alpha particle decay, the unstable nucleus emits an alpha particle reducing its proton number (atomic number) Z as well as neutron number N by 2.
XZA→YZ2A4+H24e
During β^{−}decayâ€‹, a neutron is converted to a protonâ€‹, an electron and an antineutrino. Thus, there is an increase in the atomic number.
XZA→YZ+1A+e+ν¯
During β^{+}decayâ€‹, a proton in the nucleus is converted to a neutronâ€‹, a positron and a neutrino in order to maintain the stability of the nucleus. Thus, there is a decrease in the atomic number. â€‹
XZA→YZ1A+β++ν
When a nucleus is in higher excited state or has excess of energy, it comes to the lower state in order to become stable and release energy in the form of electromagnetic radiation called gamma ray. The element in the gamma decay doesn’t change.
Therefore, alpha and beta plus decay suffer decrease in atomic number.
Question 7:
Magnetic field does not cause deflection in
(a) αrays
(b) betaplus rays
(c) betaminus rays
(d) gamma rays
Answer:
(d) gamma rays
Magnetic force acts on a charged particle, due to which it deflects from its path. The magnitude of this force is measured as
F→=q(v→×B→).
Here, q is the charge on the particle that is moving with speed v in a uniform magnetic field B.
Since alpha, betaplus and betaminus are charged particles, they suffer deflection due to the field applied. On the other hand, gamma rays are photons and due to zero charge, they do not suffer any deflection.
Question 8:
Which of the following are electromagnetic waves?
(a) αrays
(b) Betaplus rays
(c) Betaminus rays
(d) Gamma rays
Answer:
(d) Gamma rays
Alpha rays, betaplus and betaminus rays carry charged particles that show particle behaviour. On the other hand, gamma rays carry photons that show particle as well as wave behaviour. Hence, only gamma rays are electromagnetic waves.
Question 9:
Two lithium nuclei in a lithium vapour at room temperature do not combine to form a carbon nucleus because
(a) a lithium nucleus is more tightly bound than a carbon nucleus
(b) carbon nucleus is an unstable particle
(c) it is not energetically favourable
(d) Coulomb repulsion does not allow the nuclei to come very close.
Answer:
(d) Coulomb repulsion does not allow the nuclei to come very close.
Lithium atom contains 3 protons and 3 neutrons in the nucleus and 3 valence electrons. When two lithium nuclei are brought together, they repel each other. The attractive nuclear forces being shortrange are insignificant as compared to the electrostatic repulsion. Thus, the nuclei do not combine to form carbon atom because of coulomb repulsion.
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Question 10:
For nuclei with A > 100,
(a) the binding energy of the nucleus decreases on an average as A increases
(b) the binding energy per nucleon decreases on an average as A increases
(c) if the nucleus breaks into two roughly equal parts, energy is released
(d) if two nuclei fuse to form a bigger nucleus, energy is released.
Answer:
(b) the binding energy per nucleon decreases on an average as A increases
(c) if the nucleus breaks into two roughly equal parts, energy is released
Binding energy per nucleon varies in a way that it depends on the actual value of mass number (A). As the mass number (A) increases, the binding energy also increases and reaches its maximum value of 8.7 MeV for A (50−80) and for A > 100. The binding energy per nucleon decreases as A increases and the nucleus breaks into two or more atoms of roughly equal parts so as to attain stability and binding energy of mass number between 50−80.
Question 1:
Assume that the mass of a nucleus is approximately given by M = Am_{p} where A is the mass number. Estimate the density of matter in kgm^{−3} inside a nucleus. What is the specific gravity of nuclear matter?
Answer:
Given:
Mass of the nucleus, M = Am_{p}
Volume of the nucleus, V =
43πR03ADensity of the matter,
d=MV=Amp43πR03A
=3mp4×πR03=3×1.0072764×3.14(1.1)3= 3×1017 kg/m3Specific gravity of the nuclear matter =
Density of matter Density of water
∴Specific gravity =
3×1017103= 3
×10^{14} kg/m^{3}
Question 2:
A neutron star has a density equal to that of the nuclear matter. Assuming the star to be spherical, find the radius of a neutron star whose mass is 4.0 × 10^{30} kg (twice the mass of the sun).
Answer:
Given:
Mass of the neutron star, M = 4.0 × 10^{30} kg
Density of nucleus, d = 2.4
×10^{17}
Density of nucleus,
d=MVHere, V is the volume of the nucleus.
∴ V=Md=4×10302.4×1017 =10.6×1013 =16×1014If R is the radius, then the volume of the neutron star is given by
V=43πR3 ∴16×1014=43×π×R3⇒ R3=16×34×1π×1014⇒ R3=18×100π×1012∴ R=12×104×3.17 =1.585×104=15 km
Question 3:
Calculate the mass of an αparticle. Its Its binding energy is 28.2 MeV.
Answer:
Given:
Binding energy of α particle = 28.2 MeV
Let x be the mass of α particle.
We know an α particle consists of 2 protons and 2 neutrons.
Binding energy,
B=Zmp+NmnMc2Here, m_{p} = Mass of proton
m_{n} = Mass of neutron
Z= Number of protons
N = Number of neutrons
c = Speed of light
On substituting the respective values, we have
28.2 =(2×1.007276+2×1.008665 x)c2⇒ x=4.0016 u
Question 4:
How much energy is released in the following reaction:
^{7}Li + p → α + α.
Atomic mass of ^{7}Li = 7.0160 u and that of ^{4}He = 4.0026 u.
Answer:
Given:
Mass of ^{7}Li = 7.0160 u
Mass of ^{4}He = 4.0026 u.
Reaction:
Li7+p→α+α+E,Energy release
Eis given by
E=mLi7+mp2×mHe4c2=7.0160 u+1.007276 u24.0026 uc2=(8.023273 u8.0052 u) c2=0.018076 ×931 MeV =16.83 MeV
Question 5:
Find the binding energy per nucleon of
79197Au if its atomic mass is 196.96 u.
Answer:
Given:
Atomic mass of Au, A = 196.96
Atomic number of Au, Z = 79
Number of neutrons, N = 118
Binding energy,
B=(Zmp+NmnM)c2
Here, m_{p} = Mass of proton
M = Mass of nucleus
m_{n} = Mass of neutron
c = Speed of light
On substituting the respective values, we get
B=[(79×1.007276+118×1.008665) u196.96 u]c2 =198.597274196.96×931 MeV =1524.302094 MeVBinding energy per nucleon
=1524.3197=7.737 MeV
Question 6:
(a) Calculate the energy released if ^{238}U emits an αparticle. (b) Calculate the energy to be supplied to ^{238}U it two protons and two neutrons are to be emitted one by one. The atomic masses of ^{238}U, ^{234}Th and ^{4}He are 238.0508 u, 234.04363 u and 4.00260 u respectively.
Answer:
(a)
Given:
Atomic mass of ^{238}U, m(^{238}U) = 238.0508 u
Atomic mass of ^{234}Th, m(^{234}Th) = 234.04363 u
Atomic mass of ^{4}He, m(^{4}He) = 4.00260 u
When ^{238}U emits an αparticle, the reaction is given by
U238→Th234+He4Mass defect, Δm = [m(^{238}U)
(m(^{234}Th)+m(^{4}He))]
Δm = [238.0508
(234.04363 + 4.00260) = 0.00457 u
Energy released (E) when ^{238}U emits an αparticle is given by
E=∆m c2E = [0.00457 u]×931.5 MeV⇒E = 4.25467 MeV = 4.255 MeV(b)
When two protons and two neutrons are emitted one by one, the reaction will be
U233→Th234+2n+2p Mass defect, ∆m=mU238[mTh234+2mn+2mp]∆m=238.0508 u[234.04363 u+2(1.008665) u+2(1.007276) u]∆m= 0.024712 uEnergy released (E) when ^{238}U emits two protons and two neutrons is given by
E=∆mc2E=0.024712 ×931.5 MeVE=23.019=23.02 MeV
Question 7:
Find the energy liberated in the reaction
^{223}Ra → ^{209}Pb + ^{14}C.
The atomic masses needed are as follows.
^{223}Ra ^{209}Pb ^{14}C
22..018 u 208.981 u 14.003 u
Answer:
Given:
Atomic mass of ^{223}Ra, m(^{223}Ra) = 223.018 u
Atomic mass of ^{209}Pb, m(^{209}Pb) = 208.981 u
Atomic mass of ^{14}C, m(^{14}C) = 14.003 u
Reaction:
Ra223→209Pb+C14Energy, E=mR223amP209b+mC14c2 = 223.018 u208.981+14.003 u c2 =0.034×931 MeV =31.65 MeV
Question 8:
Show that the minimum energy needed to separate a proton from a nucleus with Z protons and N neutrons is
∆E=(MZ1,N+MBMZ,N)c2where M_{Z}_{,N} = mass of an atom with Z protons and N neutrons in the nucleus and M_{B} = mass of a hydrogen atom. This energy is known as protonseparation energy.
Answer:
Given:
Mass of an atom with Z protons and N neutrons = M_{Z},_{N}
Mass of hydrogen atom = M_{H}
As hydrogen contains only protons, the reaction will be given by
EZ,N→EZ1,N+p1⇒EZ,N→Ez1,N+1H1∴ Minimum energy needed to separate a proton,
∆E=(MZ1,N+MHMZ,N)c2
Question 9:
Calculate the minimum energy needed to separate a neutron from a nucleus with Z protons and N neutrons it terms of the masses M_{Z.N}‘ M_{Z}_{,N−1} and the mass of the neutron.
Answer:
Before the separation of neutron, let the mass of the nucleus be M_{Z}_{,N}.
Then, after the separation of neutron, the mass of the nucleus will be M_{Z},_{N1.}
The reaction is given by
EZ,N=EZ,N1+n01If
MNis the mass of the neutron, then the energy needed to separate the neutron b
∆Ewill be
∆E = (Final mass of nucleus + Mass of neutron − Initial mass of the nucleus)c^{2}
∆E=(MZ,N1+MNMZ,N)c2
Question 10:
^{32}P betadecays to ^{32}S. Find the sum of the energy of the antineutrino and the kinetic energy of the βparticle. Neglect the recoil of the daughter nucleus. Atomic mass of ^{32}P = 31.974 u and that of ^{32}S = 31.972 u.
Answer:
Given:
Atomic mass of ^{32}P, m(^{32}P) = 31.974 u
Atomic mass of ^{32}S, m(^{32}S) = 31.972 u
Reaction:
P32→S32+1v0+1β0Energy of antineutrino and βparticle, E = [m(^{32}P)
–m(^{32}S)]c^{2}
= (31.974 u− 31.972 u)c^{2}
= 0.002 × 931 = 1.862 MeV
Question 11:
A free neutron betadecays to a proton with a halflife of 14 minutes. (a) What is the decay constant? (b) Find the energy liberated in the process.
Answer:
Given:
Halflife period of free neutron betadecays to a proton,
T1/2= 14 minutes
Halflife period, T_{1}_{/2} =
0.6931λHere,
λ= Decay constant
∴ λ=0.69314×60 =8.25×104 s1If m_{p} is the mass of proton, let m_{n} and m_{e} be the mass of neutron and mass of electron, respectively.
∴ Energy liberated, E=[mnmp+me]c2 =[1.008665 u1.007276+0.0005486 u]c2 =0.0008404×931 MeV =782 keV
Question 12:
Complete the following decay schemes.
(a)
Ra88226→α+(b)
O819→F919+(c)
Al1325→Mg1225+
Answer:
(a) Ra88226→α24+Rn86222(b) O819→F919+e¯+v¯(c) Ar1325→Mg1225+e++v
Question 13:
In the decay ^{64}Cu → ^{64}Ni + e^{+} + v, the maximum kinetic energy carried by the positron is found to be 0.650 MeV.
(a) What is the energy of the neutrino which was emitted together with a positron of kinetic energy 0.150 MeV?
(b) What is the momentum of this neutrino in kg m s^{−1}?
Use the formula applicable to a photon.
Answer:
Given:
Maximum kinetic energy of the positron, K = 0.650 MeV
(a) Neutrino and positron are emitted simultaneously.
∴ Energy of neutrino = 0.650 − Kinetic energy of the given positron
= 0.650 − 0.150
= 0.5 MeV = 500 keV
(b) Momentum of the neutrino,
P=Ec Here, E = Energy of neutrino
c = Speed of light
⇒P=500×1.6×10193×108×103 =2.67×1022 Kgms1
Question 14:
Potassium40 can decay in three modes. It can decay by β^{−}emission, B*emission of electron capture. (a) Write the equations showing the end products. (b) Find the Qvalues in each of the three cases. Atomic masses of
Ar1840, K1940 and Ca2040are 39.9624 u, 39.9640 u and 39.9626 u respectively.
Answer:
(a) Decay of potassium40 by β^{−}emission is given by
K4019→20Ca40+β+v¯ Decay of potassium40 by β^{+} emission is given by
K4019→18Ar40+β++v Decay of potassium40 by electron capture is given by
K4019+e→18Ar40+v(b)
Q_{value} in the β^{−} decay is given by
Q_{value} = [m(_{19}K^{40}) − m(_{20}Ca^{40})]c^{2}
= [39.9640 u − 39.9626 u]c^{2}
= 0.0014
×931 MeV
= 1.3034 MeV
Q_{value} in the β^{+} decay is given by
Q_{value} = [m(_{19}K^{40}) − m(_{20}Ar^{40}) − 2m_{e}]c^{2}
= [39.9640 u − 39.9624 u − 0.0021944 u]c^{2}
= (39.9640 − 39.9624) 931 MeV − 1022 keV
= 1489.96 keV − 1022 keV
= 0.4679 MeV
Q_{value} in the electron capture is given by
Q_{value} = [ m(_{19}K^{40}) − m(_{20}Ar^{40})]c^{2}
= (39.9640 − 39.9624)uc^{2}
= 1.4890 = 1.49 MeV
Question 15:
Lithium (Z = 3) has two stable isotopes ^{6}Li and ^{7}Li. When neutrons are bombarded on lithium sample, electrons and αparticles are ejected. Write down the nuclear process taking place.
Answer:
The nuclear process taking place is shown below.
Li86+n →Li37Li37+n→Li38→Be48+v¯+eBe48→He24+He24
Question 16:
The masses of ^{11}C and ^{11}B are respectively 11.0114 u and 11.0093 u. Find the maximum energy a positron can have in the β*decay of ^{11}C to ^{11}B.
Answer:
Given:
Mass of ^{11}C, m(^{11}C) = 11.0114 u
Mass of ^{11}B, m(^{11}B) = 11.0093 u
Energy liberated in the β^{+} decay (Q) is given by
Q=mC11mB112mec2 = (11.0114 u − 11.0093 u
2
×0.0005486 u)c^{2}
= 0.0010028
×931 MeV
= 0.9336 MeV = 933.6 keV
For maximum KE of the positron, energy of neutrino can be taken as zero.
∴ Maximum KE of the positron = 933.6 keV
Question 17:
^{228}Th emits an alpha particle to reduce to ^{224}Ra. Calculate the kinetic energy of the alpha particle emitted in the following decay:
Th228→Ra224*+αRa224*→224Ra+Υ(217 keV).Atomic mass of ^{228}Th is 228.028726 u, that of ^{224}Ra is 224.020196 u and that of
H24is 4.00260 u.
Answer:
Given:
Atomic mass of ^{228}Th, m(^{228}Th) = 228.028726 u
Atomic mass of ^{224}Ra, m(^{224}Ra) = 224.020196 u
Atomic mass of
H24, m(
H24) = 4.00260 u
Mass of ^{224}Ra = 224.020196 × 931 + 0.217 MeV = 208563.0195 MeV
Kinetic energy of alpha particle, K =
mTh228mRa224+mH24c^{2}
= (228.028726 × 931) − [(208563.0195 + 4.00260 × 931]
= 5.30383 MeV = 5.304 MeV
Question 18:
Calculate the maximum kinetic energy of the beta particle emitted in the following decay scheme:
^{12}N → ^{12}C* + e^{+} + v
^{12}C* → ^{12}C + γ (4.43MeV).
The atomic mass of ^{12}N is 12.018613 u.
Answer:
Given:
Atomic mass of ^{12}N, m(^{12}N) = 12.018613 u
^{12}N → ^{12}C* + e^{+} + v
^{12}C* → ^{12}C + γ (4.43 MeV)
Net reaction is given by
^{ }^{12}N → ^{12}C + e^{+} + v + γ (4.43 MeV)
Q_{value} of the
β+decay will be
Q_{value}= [ m(^{12}N)
(m(^{12}C*) + 2m_{e})]c^{2}
= [12.018613
×931 MeV
(12
×931 + 4.43) MeV
(2
×511) keV]
= [11189.3287
11176.43
1.022] MeV
= 11.8767 MeV = 11.88 MeV
The maximum kinetic energy of beta particle will be 11.88 MeV, assuming that neutrinos have zero energy.
Question 19:
The decay constant of
Hg80197(electron capture to
Au79197) is 1.8 × 10^{−4} S^{−1}. (a) What is the halflife? (b) What is the averagelife? (c) How much time will it take to convert 25% of this isotope of mercury into gold?
Answer:
Given:
Decay constant of
Hg80197,
λ= 1.8 × 10^{−4} s
1
(a)
Halflife,
T12=0.693λ
⇒T1/2=0.6931.8×104 = 3850 s=64 minutes
(b)
Average life, Tav=T1/20.693 =640.693 =92 minutes(c)
Number of active nuclei of mercury at t = 0
= N_{0}
= 100
Active nuclei of mercury left after conversion of 25% isotope of mercury into gold = N = 75
Now,
NN0=eλtHere, N = Number of inactive nuclei
N_{0} = Number of nuclei at t = 0
λ= Disintegration constant
On substituting the values, we get
75100=eλt
⇒0.75=eλx⇒ In 0.75=λt⇒ t=In 0.750.00018 =1600 s
Page No 443:
Question 20:
The halflife of ^{199}Au is 2.7 days. (a) Find the activity of a sample containing 1.00 µg of ^{198}Au. (b) What will be the activity after 7 days? Take the atomic weight of ^{198}Au to be 198 g mol^{−1}.
Answer:
Given:
Halflife of ^{199}Au, T_{1}_{/2}= 2.7 days
Disintegration constant,
λ=0.693T1/2=
0.6932.7×24×60×60=2.97×106 s1Number of atoms left undecayed, N =
1×106×6.023×1023198Now, activity,
A0=λN
=
1×106×6.023×1023198×2.97×106 = 0.244 Ci
(b) After 7 days,
Activity,
A=A0eλtHere, A_{0} = 0.244 Ci
∴ A =0.244×e2.97×106×7×3600×24 =0.244×e17962.56×104 =0.040 Ci
Question 21:
Radioactive ^{131}I has a halflife of 8.0 days. A sample containing ^{131}I has activity 20 µCi at t = 0. (a) What is its activity at t = 4 days? (b) What is its decay constant at t = 4.0 days?
Answer:
Given:
Halflife of radioactive ^{131}I, T_{1}_{/2} = 8 days
Activity of the sample at t = 0, A_{0} = 20 µCi
Time, t = 4 days
(a) Disintegration constant,
λ=0.693T1/2=0.0866Activity (A) at t = 4 days is given by
A=A0eλt⇒A=20×106×e0.0866×4 =1.41×106 Ci=14 μCi(b) Decay constant is a constant for a radioactive sample and depends only on its halflife.
λ=0.693[8×24×3600] =1.0026×106 s1
Question 22:
The decay constant of ^{238}U is 4.9 × 10^{−18} S^{−1}. (a) What is the averagelife of ^{238}U? (b) What is the halflife of ^{238}U? (c) By what factor does the activity of a ^{238}U sample decrease in 9 × 10^{9} years?
Answer:
Given:
Decay constant,
λ= 4.9 × 10^{−18} s^{−1}
(a) Average life of uranium
τis given by
τ=1λ =14.9×1018 =14.9×1018 s =10164.9×365×24×36 years =10164.9×365×24×36 years = 6.47×107×1016 years =6.47×109 years(b) Halflife of uranium
T12is given by
T12=0.693λ=0.6934.9×1018 =0.6934.9×1018 s =0.1414×1018 s =0.1414×1018365×24×3600 =1414×1012365×24×36 = 4.48×103×1012 =4.5×109 years(c) Time, t = 9 × 10^{9} years
Activity (A) of the sample, at any time t, is given by
A=A02tT1/2 Here, A_{0} = Activity of the sample at t = 0
∴ A0A=29×1094.5×109=22=4
Question 23:
A certain sample of a radioactive material decays at the rate of 500 per second at a certain time. The count rate falls to 200 per second after 50 minutes. (a) What is the decay constant of the sample? (b) What is its halflife?
Answer:
Given:
Initial rate of decay, A_{0} = 500,
Rate of decay after 50 minutes, A = 200
Time, t = 50 min
= 50
×60
= 3000 s
(a)
Activity, A = A_{0}e^{−λ}^{t}
Here,
λ= Disintegration constant
∴ 200 = 500 × e^{−50}^{×60×λ}
⇒25=e3000λ⇒ In25=3000λ⇒ λ=3.05×104 s1(b)
Halflife, T1/2=0.693λ =2272.13 s =38 min
Question 24:
The count rate from a radioactive sample falls from 4.0 × 10^{6} per second to 1.0 × 10^{6} per second in 20 hours. What will be the count rate 100 hours after the beginning?
Answer:
Given:
Initial count rate of radioactive sample, A_{0} = 4 × 10^{6} disintegration/sec
Count rate of radioactive sample after 20 hours, A = 1 × 10^{6} disintegration/sec
Time, t = 20 hours
Activity of radioactive sample
Ais given by
A=A02tT2Here, T1/2=Halflife periodOn substituting the values of A0 and A, we have 2tT2=22⇒tT2=2⇒T2=t/2=20 h/2=10 h100 hours after the beginning,
Count rate, A”=A02tT2⇒A”=4×1062100/10 =0.390625×104 =3.9×103 disintegrations/sec
Question 25:
The halflife of ^{226}Ra is 1602 y. Calculate the activity of 0.1 g of RaCl_{2} in which all the radium is in the form of ^{226}Ra. Taken atomic weight of Ra to be 226 g mol^{−1} and that of Cl to be 35.5 g mol^{−1}.
Answer:
Given:
Halflife of radium, T_{1}_{/2} = 1602 years
Atomic weight of radium = 226 g/mole
Atomic weight of chlorine = 35.5 g/mole
Now,
1 mole of RaCl_{2} = 226 + 71 = 297 g
297 g = 1 mole of RaCl_{2}
0.1 g =
1297×0.1 mole of RaCl2Total number of atoms in 0.1 g of RaCl_{2}, N
=0.1×6.023×1023297 = 0.02027×1022
∴ No of atoms, N = 0.02027
×10^{22}
Disintegration constant, λ=0.693T12 =0.6931602×365×24×3600 =1.371×1011Activity of radioactive sample, A =
λN =
1.371×1011×2.027×1020 =
2.8×109 disintegrations/second
Question 26:
The halflife of a radioisotope is 10 h. Find the total number of disintegration in the tenth hour measured from a time when the activity was 1 Ci.
Answer:
Given:
Halflife of radioisotope,
T1/2= 10 hrs
Initial activity, A_{0} = 1 Ci
Disintegration constant,
λ=0.69310×3600 s1Activity of radioactive sample,
A=A0eλt
Here, A_{0} = Initial activity
λ= Disintegration constant
t = Time taken
After 9 hours,
Activity,
A=A0eλt=1×e0.69310×3600×9=0.536 Ci∴ Number of atoms left, N =
Aλ=
0.536×10×3.7×1010×36000.693=103.023×1013After 10 hrs,
Activity, A”=A0eλt =1×e0.69310×10=0.5 CiNumber of atoms left after the 10^{th} hour
N”will be
A”=λN”N”=A”λ =0.5×3.7×1010×3.6000.693/10 =26.37×1010×3600=96.103×1013Number of disintegrations = (103.023 − 96.103) × 10^{13}
= 6.92 × 10^{13}
Question 27:
The selling rate of a radioactive isotope is decided by its activity. What will be the secondhand rate of a one month old ^{32}P(t_{1}_{/2} = 14.3 days) source if it was originally purchased for 800 rupees?
Answer:
Given:
Halflife of ^{32}P source,
T12= 14.3 days
Time, t = 30 days = 1 month
Here, the selling rate of a radioactive isotope is decided by its activity.
∴ Selling rate = Activity of the radioactive isotope after 1 month
Initial activity, A_{0} = 800 disintegration/sec
Disintegration constant
λis given by
λ=0.693T12=
0.69314.3 days1Activity
Ais given by
A = A_{0}e^{−}^{λt}
Here,
λ= Disintegration constant
∴Activity of the radioactive isotope after one month (selling rate of the radioactive isotope)
Ais given below.
A=800×e0.69314.3×30 =800×0.233669 =186.935=Rs 187
Question 28:
^{57}Co decays to ^{57}Fe by β^{+}– emission. The resulting ^{57}Fe is in its excited state and comes to the ground state by emitting γrays. The halflife of β^{+}– decay is 270 days and that of the γemissions is 10^{−8} s. A sample of ^{57}Co gives 5.0 × 10^{9} gamma rays per second. How much time will elapse before the emission rate of gamma rays drops to 2.5 × 10^{9} per second?
Answer:
According to the question, when the β^{+} decays to half of its original amount, the emission rate of γrays will drop to half. For this, the sample will take 270 days.
Therefore, the required time is 270 days.
Question 29:
Carbon (Z = 6) with mass number 11 decays to boron (Z = 5). (a) Is it a β^{+}decay or a β^{−}decay? (b) The halflife of the decay scheme is 20.3 minutes. How much time will elapse before a mixture of 90% carbon11 and 10% boron11 (by the number of atoms) converts itself into a mixture of 10% carbon11 and 90% boron11?
Answer:
(a) The reaction is given by
C6→B5+β++v It is a β^{+} decay since atomic number is reduced by 1.
(b) Halflife of the decay scheme, T
_{1/2 } = 20.3 minutesDisintegration constant,
λ=0.693T12=
0.69320.3 min1If t is the time taken by the mixture in converting, let the total no. of atoms be 100N_{0}_{.}
Carbon  Boron  
Initial  90 N_{0}  10 N_{0} 
Final  10 N_{0}  90 N_{0} 
N = N_{0}e^{−λ}^{t}
Here, N_{0} = Initial number of atoms
N = Number of atoms left undecayed
10N_{0} = 90N_{0}e^{−λ}^{t} ( For carbon)
⇒ 19=e0.69320.3×t ⇒ In19=0.69320.3t⇒ t= 64.36=64 min
Question 30:
4 × 10^{23} tritium atoms are contained in a vessel. The halflife of decay tritium nuclei is 12.3 y. Find (a) the activity of the sample, (b) the number of decay in the next 10 hours (c) the number of decays in the next 6.15 y.
Answer:
Given:
Number of tritium atoms, N_{0} = 4 × 10^{23}
Halflife of tritium nuclei,
T12= 12.3 years
Disintegration constant,
λ=0.693T12=0.69312.3years
1
Activity of the sample
Ais given by
A_{0} =
dNdt=
λN0
⇒A0=0.693t1/2N0 =0.69312.3×4×1023 disintegration/year =0.693×4×102312.3×3600×24×365 disintegration/sec =7.146×1014 disintegration/sec(b) Activity of the sample, A = 7.146
×10^{14} disintegration/sec
Number of decays in the next 10 hours=7.146×1014×10×3600 =257.256×1017 =2.57×1019(c) Number of atoms left undecayed,
N=N0eλt Here, N_{0} = Initial number of atoms
∴ N=4×1023×e0.69312.3×6.15=2.83×1023
Number of atoms disintegrated = (N_{0}
–N) = (4
2.83) × 10^{23} = 1.17 × 10^{23}
Question 31:
A point source emitting alpha particles is placed at a distance of 1 m from a counter which records any alpha particle falling on its 1 cm^{2} window. If the source contains 6.0 × 10^{16} active nuclei and the counter records a rate of 50000 counts/second, find the decay constant. Assume that the source emits alpha particles uniformly in all directions and the alpha particles fall nearly normally on the window.
Answer:
Given:
Counts received per second = 50000 Counts/second
Number of active nuclei, N = 6 × 10^{16}
Total counts radiated from the source,
dNdt= Total surface area × 50000 counts/cm^{2}
= 4 × 3.14 × 1 × 10^{4} × 5 × 10^{4}
= 6.28 × 10^{9} Counts
We know
dNdt=λNHere, λ = Disintegration constant
∴ λ=6.28×1096×1016 =1.0467×107 =1.05×107s1
Question 32:
^{238}U decays to ^{206}Pb with a halflife of 4.47 × 10^{9} y. This happens in a number of steps. Can you justify a single half for this chain of processes? A sample of rock is found to contain 2.00 mg of ^{238}U and 0.600 mg of ^{206}Pb. Assuming that all the lead has come from uranium, find the life of the rock.
Answer:
Given:
Halflife of ^{238}U, t_{1}_{/2} = 4.47 × 10^{9} years
Total number of atoms present in the rock initially, N0=6.023×1023×2238+6.023×1023×0.6206 =12.046238+3.62206×1020 =0.0505+0.0175×1020 =0.0680×1020Now, N = N_{0}e
λt
Here,
λ= Disintegration constant
t = Life of the rock
⇒ N=N0e0.693t1/2×t⇒ 0.0505=0.0680e0.6934.47×109×t⇒ ln0.05050.0680=0.69314.47×109×t⇒t=1.92×109 years
Question 33:
When charcoal is prepared from a living tree, it shows a disintegration rate of 15.3 disintegrations of ^{14}C per gram per minute. A sample from an ancient piece of charcoal shows ^{14}C activity to be 12.3 disintegrations per gram per minute. How old is this sample? Halflife of ^{14}C is 5730 y.
Answer:
Given:
Initial activity of charcoal, A_{0} = 15.3 disintegrations per gram per minute
Halflife of charcoal,
T12= 5730 years
Final activity of charcoal after a few years, A = 12.3 disintegrations per gram per minute
Disintegration constant,
λ=0.693T12=
0.6935370 y1Let the sample take a time of t years for the activity to reach 12.3 disintegrations per gram per minute.
Activity of the sample,
A=A0eλt
A=A0e0.6935730×t⇒In12.315.3=0.6935730t⇒0.218253=0.6935730×t⇒t=1804.3 years
Question 34:
Natural water contains a small amount of tritium (
H13). This isotope betadecays with a halflife of 12.5 years. A mountaineer while climbing towards a difficult peak finds debris of some earlier unsuccessful attempt. Among other things he finds a sealed bottled of whisky. On returning, he analyses the whisky and finds that it contains only 1.5 per cent of the
H13radioactivity as compared to a recently purchased bottle marked ‘8 years old’. Estimate the time of that unsuccessful attempt.
Answer:
Given:
Halflife time of tritium,
T12= 12.5 years
Disintegration constant,
λ=0.69312.5 per yearLet A_{0} be the activity, when the bottle was manufactured.
Activity after 8 years
Ais given by
A=A0e0.69312.5×8 …(1)
Let us consider that the mountaineering had taken place t years ago.
Then, activity of the bottle
A’on the mountain is given by
A’=A0eλtHere, A’ = (Activity of the bottle manufactured 8 years ago) × 1.5%
A’=A0e0.69312.5×8×0.015 …(2)Comparing (1) and (2)
0.69312.5 t=0.6931×812.5+In [0.015]⇒ 0.69312.5 t=0.69312.5×84.1997⇒0.693 t=58.040⇒ t=83.75 years
Question 35:
The count rate of nuclear radiation coming from a radiation coming from a radioactive sample containing ^{128}I varies with time as follows.
Time t (minute):  0  25  50  75  100 
Ctount rate R (10^{9} s^{−1}):  30  16  8.0  3.8  2.0 
(a) Plot In (R_{0}/R) against t. (b) From the slope of the best straight line through the points, find the decay constant λ. (c) Calculate the halflife t_{1}_{/2}.
Answer:
(a) For t = 0,
lnR0R=In30×10930×109=0For t = 25 s,
lnR0R2=In30×10916×109=0.63For t = 50 s,
InR0R3=In30×1098×109=1.35For t = 75 s,
lnR0R4=In30×1093.8×109=2.06For t = 100 s,
InR0R5=In30×1092×109=2.7The required graph is shown below.
(b) Slope of the graph = 0.028
∴ Decay constant,
λ= 0.028 min
1
The halflife period
T12is given by
T12=0.693λ =0.6930.028=25 min
Question 36:
The halflife of ^{40}K is 1.30 × 10^{9} y. A sample of 1.00 g of pure KCI gives 160 counts s^{−1}. Calculate the relative abundance of ^{40}K (fraction of ^{40}K present) in natural potassium.
Answer:
Given:
Halflife period of ^{40}K,
T12= 1.30 × 10^{9} years
Count given by 1 g of pure KCI, A = 160 counts/s
Disintegration constant,
λ=0.693T12Now, activity, A = λN
⇒160=0.693t1/2 ×N⇒ 160=0.6931.30×109×365×86400× N⇒ N=160×1.30×365×86400×1090.693⇒N=9.5×1018
6.023 × 10^{23} atoms are present in 40 gm.
Thus, 9.5 × 1018 atoms will be present in40×9.5×10186.023×1023 gm.=4×9.5×1046.023 gm=6.309×104=0.00063 gmRelative abundance of ^{40}K in natural potassium = (2 × 0.00063 × 100)% = 0.12%
Question 37:
Hg80197decay to
Au79197through electron capture with a decay constant of 0.257 per day. (a) What other particle or particles are emitted in the decay? (b) Assume that the electron is captured from the K shell. Use Moseley’s law √v = a(Z − b) with a = 4.95 × 10^{7} s^{−1}^{/2} and b = 1 to find the wavelength of the K_{α} Xray emitted following the electron capture.
Answer:
Given:
Decay constant of electron capture = 0.257 per day
(a) The reaction is given as
Hg80197+e→Au79197+vThe other particle emitted in this reaction is neutrino v.
(b) Moseley’s law is given by
v = a(Z − b)
We know
v=cλHere, c = Speed of light
λ= Wavelength of the K_{α} Xray
cλ=4.95×107(791) =4.95×107×78⇒cλ=(4.95×78)2×1014⇒λ=3×108149073.21×1014 =20 pm
Question 38:
A radioactive isotope is being produced at a constant rate dN/dt = R in an experiment. The isotope has a halflife t_{1}_{/2}. Show that after a time t >> t_{1}_{/2} the number of active nuclei will become constant. Find the value of this constant.
Answer:
Given:
Half life period of isotope = t_{1}_{/2}
Disintegration constant,
λ=0.693t1/2Rate of Radio active decay
Ris given by,
R=dNdtWe are to show that after time t >> t_{1}_{/2} the number of active nuclei is constant.
dNdtPresent=R=dNdtdecay∴ R = dNdtdecayRate of radioactive decay,
R=λNHere, λ = Radioactive decay constant
N = Constant number
R=0.693t1/2×N⇒Rt1/2=0.693N⇒N=Rt1/20.693This value of N should be constant.
Question 39:
Consider the situation of the previous problem. Suppose the production of the radioactive isotope starts at t = 0. Find the number of active nuclei at time t.
Answer:
Let the number of atoms present at t = 0 be N_{0}.
Let N be the number of radioactive isotopes present at time t.
Then,
N = N_{0}e^{−λt}
Here,
λ= Disintegration constant
∴ Number of radioactive isotopes decayed = N_{0} − N = N_{0} − N_{0}e^{−}^{λt}
= N_{0} (1−e^{−}^{λt}) …(1)
Rate of decay
Ris given by
R = λN_{0} …(2)
Substituting the value of N_{0} from equation (2) to equation (1), we get
N=N0(1eλt) =Rλ(1eλt)
Question 40:
In an agricultural experiment, a solution containing 1 mole of a radioactive material (t_{1}_{/2} = 14.3 days) was injected into the roots of a plant. The plant was allowed 70 hours to settle down and then activity was measured in its fruit. If the activity measured was 1 µCi, what per cent of activity is transmitted from the root to the fruit in steady state?
Answer:
Given:
Initial no of atoms, N_{0} = 1 mole = 6 × 10^{23} atoms
Halflife of the radioactive material, T_{1}_{/2} = 14.3 days
Time taken by the plant to settle down, t = 70 h
Disintegration constant,
λ=0.693t1/2=
0.69314.3×24h
1
N = N_{0}e^{−λ}^{t}
=6×1023×e0.693×7014.3×24=6×1023×0.868=5.209×1023
Activity, R=dNdt=5.209×1023×0.69314.3×24 =0.0105×10233600dis/hr =2.9×106×1023 dis/sec =2.9×1017dis/secFraction of activity transmitted = 1 μCi2.9×1017×100% =1×3.7×1042.9×1017×100% =1.275×1011%
Question 41:
A vessel of volume 125 cm^{3} contains tritium (^{3}H, t_{1}_{/2} = 12.3 y) at 500 kPa and 300 K. Calculate the activity of the gas.
Answer:
Given:
Volume of the vessel, V = 125 cm^{3} = 0.125 L
Halflife time of tritium, t_{1}_{/2} = 12.3 y = 3.82 × 10^{8} s
Pressure, P = 500 kpa = 5 atm
Temperature, T = 300 K,
Disintegration constant,
λ=0.693t1/2 =
0.6933.82×108=0.1814×108=1.81×109 s1No. of atoms left undecayed, N = n × 6.023 × 10^{23}
=5×0.1258.2×102×3×102×6.023×1023 ∵ n=PVRT=1.5×1022 atomsActivity, A = λN
= 1.81 × 10^{−9} × 1.5 × 10^{22} = 2.7 × 10^{13} disintegration/sec
∴A=2.7×10133.7×1010 Ci=729.72 Ci
Page No 444:
Question 42:
Bi83212can disintegrate either by emitting an αparticle of by emitting a β^{−}particle. (a) Write the two equations showing the products of the decays. (b) The probabilities of disintegration αand βdecays are in the ratio 7/13. The overall halflife of ^{212}Bi is one hour. If 1 g of pure ^{212}Bi is taken at 12.00 noon, what will be the composition of this sample at 1 P.m. the same day?
Answer:
Given:
Halflife of ^{212}Bi, T_{1}_{/2} = 1 h
1
When
Bi83212disintegrates by emitting an αparticle
Bi83212→T812081+He24(α)When
Bi83212disintegrates by emitting a β^{−}particle
Bi83212→P084212+β+v¯Halflife period of ^{212}Bi,
T12= 1 h
1
At t = 0, the amount of ^{212}Bi present = 1 g
At t = 1 = One halflife,
Amount of ^{212}Bi present = 0.5 g
Probability of disintegration of αdecay and βdecay are in the ratio
713.
In 20 g of ^{212}Bi, the amount of ^{208}Ti formed = 7 g
In 1 g of ^{212}Bi, the amount of ^{208}Ti formed = 7/20 g
∴ Amount of ^{208}Ti present in 0.5 g =
720×0.5=0.175 gIn 20 g of ^{212}Bi, the amount of ^{212}Po formed = 13 g
In 1 g of ^{212}Bi, the amount of ^{212}Po formed = 13/20 g
∴ Amount of ^{212}Po present in 0.5 g =
1320×0.5=0.325 g
Question 43:
A sample contains a mixture of ^{108}Ag and ^{110}Ag isotopes each having an activity of 8.0 × 10^{8} disintegration per second. ^{110}Ag is known to have larger halflife than ^{108}Ag. The activity A is measured as a function of time and the following data are obtained.
Time (s)

Activity (A) (10^{8} disinte grations s^{−1}) 
Time (s)

Activity (A 10^{8} disintegrations s^{−1}) 
20 40 60 80 100 
11.799 9.1680 7.4492 6.2684 5.4115 
200 300 400 500 
3.0828 1.8899 1.1671 0.7212 
(a) Plot ln (A/A_{0}) versus time. (b) See that for large values of time, the plot is nearly linear. Deduce the halflife of ^{110}Ag from this portion of the plot. (c) Use the halflife of ^{110}Ag to calculate the activity corresponding to ^{108}Ag in the first 50 s. (d) Plot In (A/A_{0}) versus time for ^{108}Ag for the first 50 s. (e) Find the halflife of ^{108}Ag.
Answer:
(a) Activity, A_{0} = 8 × 10^{8} dis/sec
(i)
InA1A0=In11.7948=0.389(ii)
InA2A0=In9.16808=0.12362(iii)
InA3A0=In7.44928=0.072(iv)
InA4A0=In6.26846=0.244(v)
InA5A=
In5.41158=0.391(vi)
InA6A0=
In3.08288=0.954(vii)
InA7A0=
In91.88998=1.443(viii)
In1.16718=In90.72128=1.93(ix)
In0.72128=In90.72128=2.406The required graph is given below.
(b) Halflife of ^{110}Ag = 24.4 s
(c) Halflife of ^{110}Ag,
T12= 24.4 s
Decay constant,
λ=0.693T12
⇒
λ=0.69324.4=0.0284∴ t = 50 sec
Activity, A=A0eλt =8×108×e0.0284×50 =1.93×108(d)
(e) The halflife period of ^{108}Ag that you can easily watch in your graph is 144 s.
Question 44:
A human body excretes (removes by waste discharge, sweating, etc.) certain materials by a law similar to radioactivity. If technetium is injected in some form in a human body, the body excretes half the amount in 24 hours. A patient is given an injection containing ^{99}Tc. This isotope is radioactive with a halflife of 6 hours. The activity from the body just after the injection is 6 μCi. How much time will elapse before the activity falls to 3 μCi?
Answer:
Given:
Time taken by the body to excrete half the amount, t_{1} = 24 hours
Halflife of radioactive isotope, t_{2} = 6 hours
Initial activity, A_{0} = 6 μCi
Let after time t, activity of the sample be A.
Halflife period
T1/2is given by
T1/2=t1t2t1+t2=24×624+6 =24×630=4.8 hActivity
Aat time t is given by
∴ A= A02t/T1/2⇒ 3μCi=6μCi2t/4.8⇒6μCi2t/4.8=3⇒ t=4.8 h
Question 45:
A charged capacitor of capacitance C is discharged through a resistance R. A radioactive sample decays with an averagelife τ. Find the value of R for which the ratio of the electrostatic field energy stored in the capacitor to the activity of the radioactive sample remains constant in time.
Answer:
Discharging of a capacitor through a resistance R is given by
Q=qet/CRHere, Q = Charge left
q = Initial charge
C = Capacitance
R = Resistance
Energy, E =
12Q2C=
q2e2t/CR2CActivity, A = A_{0}e
λt
Here, A_{0} = Initial activity
λ= Disintegration constant
∴ Ratio of the energy to the activity =
EA=q2×e2t/CR2CA0eλtSince the terms are independent of time, their coefficients can be equated.
2tCR=λt
⇒λ=2CR
⇒1τ=2CR
⇒R=2τC
Question 46:
Radioactive isotopes are produced in a nuclear physics experiment at a constant rate dN/dt = R. An inductor of inductance 100 mH, a resistor of resistance 100 Ω and a battery are connected to form a series circuit. The circuit is switched on at the instant the production of radioactive isotope starts. It is found that i/N remains constant in time where i is the current in the circuit at time t and N is the number of active nuclei at time t. Find the halflife of the isotope.
Answer:
Given:
Resistance of resistor, R = 100 Ω
Inductance of an inductor, L = 100 mH
Current
iat any time
tis given by
i=i01eRtLNumber of active nuclei
Nat any time
tis given by
N=N0eλtWhere N_{0} = Total number of nuclei
λ= Disintegration constant
Now,
iN=i01etR/LN0eλtAs
iNis independent of time, coefficients of t are equal.
Let
t12be the halflife of the isotope.
⇒RL=λ⇒RL=0.693t12⇒t12=0.693×103=6.93×104 s
Question 47:
Calculate the energy released by 1g of natural uranium assuming 200 MeV is released in each fission event and that the fissionable isotope ^{235}U has an abundance of 0.7% by weight in natural uranium.
Answer:
235 g of uranium contains 6.02 × 10^{23} atoms.
1 g of uranium =
1235×6.023×1023atoms
∴ 0.7 g of uranium =
1235×6.023×1023 ×0.007atoms
1 atom gives 200 MeV.
∴ Total energy released =
6.023×1023×0.007×200×106×1.6×1019235 J= 5.74
×10^{8 }J
Question 48:
A uranium reactor develops thermal energy at a rate of 300 MW. Calculate the amount of ^{235}U being consumed every second. Average released per fission is 200 MeV.
Answer:
Given:
Rate of development of thermal energy = 300 MW
Average energy released per fission = 200 MeV
Let N be the number of atoms disintegrating per second.
Then, the total energy emitted per second will be
N
×200
×10^{6}
×1.6
×10
19= Power
N
×200
×10^{6}
×1.6
×10
19= 300
×10^{6}
⇒N=32×1.6×1019 = 33.2×1019 atoms
6.023 × 10^{23} atoms = 238 gm of U^{235}
33.2×1019 atoms will present in 238×3×10196×1023×3.2=3.7 mg
Question 49:
A town has a population of 1 million. The average electric power needed per person is 300 W. A reactor is to be designed to supply power to this town. The efficiency with which thermal power is converted into electric power is aimed at 25%. (a) Assuming 200 MeV to thermal energy to come from each fission event on an average, find the number of events that should take place every day. (b) Assuming the fission to take place largely through ^{235}U, at what rate will the amount of ^{235}U decrease? Express your answer in kg per day. (c) Assuming that uranium enriched to 3% in ^{235}U will be used, how much uranium is needed per month (30 days)?
Answer:
(a) Total population of the town = 1 million = 10^{6 }
Average electric power needed per person = 300 W
Total power used by the town in one day = 300 × 10^{6} × 60 × 60 × 24 J = 300 × 86400 ×10^{6} J
Energy generated in one fission = 200 × 10^{6} × 1.6 × 10^{−19} J =3.2 × 10^{−11} J
The efficiency with which thermal power is converted into electric power is 25%.
Therefore, Electrical energy is given by
∴Electrical energy, E=3.2×1011×25100E=8×1012 JLet the number of fission be N.
So, total energy of N fissions = N × 8 × 10^{−12}
As per the question,
N × 8 × 10^{−12} = 300 × 86400 × 10^{6} J
N = 3.24 × 10^{24}
(b) Number of moles required per day n =
N6.023×1023
⇒n = 3.24×10246.023×1023 = 5.38 molSo, the amount of uranium required per day = 5.38 × 235
= 1264.3 gm = 1.2643 kg
(c) Total uranium needed per month = 1.264 × 30 kg
Let x kg of uranium enriched to 3% in ^{235}U be used.
⇒ x×3100 = 1.264×30⇒x = 1264 kg
Question 50:
Calculate the Qvalues of the following fusion reactions:
(a)
H12+H12→H13+H11(b)
H12+H12→He23+n(c)
H12+H13→He24+n.
Atomic masses are
m(H12)=2.014102u,m(H13)=3.016049u,m(H23e)=3.016029u,m(H24e)=4.002603u.
Answer:
(a) Q=2×m H12mH33+mH13c2 =(4.0282044.023874)×931 MeV =4.05 MeV(b) Q=2×m H12(m H23+mn)c2 =[4.0282044.024694)×931 =0.00351×931 =3.25 MeV(c) Q=m H12+m H13m He14mnc2 =(2.014102+3.0160494.0026031.008665)×931 =17.57 MeV
Question 51:
Consider the fusion in helium plasma. Find the temperature at which the average thermal energy 1.5 kT equals the Coulomb potential energy at 2 fm.
Answer:
Given:
Average thermal energy, E = 1.5 kT
Point of coulomb potential energy = 2 fm
Potential energy is given by
U =
Kq1q2r …..(1)
Here, K =
14πε0= 9
×10^{9}
Charge,
q1=q2=2×1.6×1019 C
Average kinetic energy
Eis given by
E=32kT …..(2)
Here, k = Boltzman constant
T = Temperature
Equating equation (1) and (2), we get
kq1q2r=32kT
⇒T=2Kq1q23kr =
2×9×109×4×1.6×101923×1.38×1023×2×1015 = 2.23
×10^{10} K
Question 52:
Calculate the Qvalue of the fusion reaction
^{4}He + ^{4}He = ^{8}Be.
Is such a fusion energetically favourable? Atomic mass of ^{8}Be is 8.0053 u and that of ^{4}He is 4.0026 u.
Answer:
Given:
Atomic mass of ^{8}Be = 8.0053 u
Atomic mass of ^{4}He = 4.0026 u
Required Qvalue=(2×4.00268.0053) c2 =0.0001×931 MeV =0.0001×931×106 eV =93.1 KeVNo, such reaction is not favourable.
Question 53:
Calculate the energy that can be obtained from 1 kg of water through the fusion reaction
^{2}H + ^{2}H → ^{3}H + p.
Assume that 1.5 × 10^{−2}% of natural water is heavy water D_{2}O (by number of molecules) and all the deuterium is used for fusion.
Answer:
18 g of water contains 6.023
×10^{23 }molecules.
∴ 1000 g of water =
6.023×1023×100018=3.346×1025molecules
% of deuterium =
3.346×1025 × 0.015100= 0.05019
×10^{23}
Energy of deuterium =
30.4486×1025
=2×mH2mH3mpc2=2×2.014102 u3.016049 u1.007276 uc2=0.004879×931 MeV=4.542349 MeV=7.262 ×1013 J Total energy = 0.05019
×10^{23}
×7.262
×10
13 J
= 3644 MJ
HC Verma Solutions for Class 11 Physics – Part 1
 Chapter 1 – Introduction to Physics
 Chapter 2 – Physics and Mathematics
 Chapter 3 – Rest and Motion: Kinematics
 Chapter 4 – The Forces
 Chapter 5 – Newton’s Laws of Motion
 Chapter 6 – Friction
 Chapter 7 – Circular Motion
 Chapter 8 – Work and Energy
 Chapter 9 – Center of Mass, Linear Momentum, Collision
 Chapter 10 – Rotational Mechanics
 Chapter 11 – Gravitation
 Chapter 12 – Simple Harmonic Motion
 Chapter 13 – Fluid Mechanics
 Chapter 14 – Some Mechanical Properties of Matter
 Chapter 15 – Wave Motion and Wave on a String
 Chapter 16 – Sound Wave
 Chapter 17 – Light Waves
 Chapter 18 – Geometrical Optics
 Chapter 19 – Optical Instruments
 Chapter 20 – Dispersion and Spectra
 Chapter 21 – Speed of Light
 Chapter 22 – Photometry
HC Verma Solutions for Class 12 Physics – Part 2
 Chapter 23 – Heat and Temperature
 Chapter 24 – Kinetic Theory of Gases
 Chapter 25 – Calorimetry
 Chapter 26 – Laws of Thermodynamics
 Chapter 27 – Specific Heat Capacities of Gases
 Chapter 28 – Heat Transfer
 Chapter 29 – Electric Field and Potential
 Chapter 30 – Gauss’s Law
 Chapter 31 – Capacitors
 Chapter 32 – Electric Current in Conductors
 Chapter 33 – Thermal and Chemical Effects of Electric Current
 Chapter 34 – Magnetic Field
 Chapter 35 – Magnetic Field due to a Current
 Chapter 36 – Permanent Magnets
 Chapter 37 – Magnetic Properties of Matter
 Chapter 38 – Electromagnetic Induction
 Chapter 39 – Alternating Current
 Chapter 40 – Electromagnetic Waves
 Chapter 41 – Electric Current through Gases
 Chapter 42 – Photoelectric Effect and Wave Particle Duality
 Chapter 43 – Bohr’s Model and Physics of the Atom
 Chapter 44 – Xrays
 Chapter 45 – Semiconductors and Semiconductor Devices
 Chapter 46 – The Nucleus
 Chapter 47 – The Special Theory of Relativity