HC Verma Solutions for Class 12 Physics 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 proton-proton, neutron-proton and neutron-neutron 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 short-ranged (a few fm), strong and attractive forces. They also exert electrostatic repulsive force (long-ranged). 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 break-off 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 24Mg nucleus or two 12C nuclei. In which of the two cases more energy will be liberated?

Answer:

If we assemble 6 protons and 6 neutrons to form 12C nucleus, 92.15 MeV (product of mass number and binding energy per nucleon of carbon-12) 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 24Mg atom, 198.25 MeV of energy (binding energy) is released. Hence, in case of â€‹24Mg 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 break-off its constituent nucleons as energy is related to mass by the relation given below.
E = Δmc2

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 α-particle 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 α-particle 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 middle-weight 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 12C atom, 14C 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

12C and 14C 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€‹12​C has 6 protons, 6 electrons and 6 neutrons, whereas â€‹14C 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 1H​1), only protons are present. Due to this, the mass number is equal to the atomic number.

Question 5:

The graph of ln(R/R0) 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 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.  neglect gravitational force. When the separation is 1 fm.
(a) Fpp > Fpn = Fnn
(b) Fpp = Fpn = Fnn
(c) Fpp > Fpn > Fnn
(d) Fpp < Fpn = Fnn

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 Fpp, Fpn and Fnn 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) Fpp > Fpn = Fnn
(b) Fpp = Fpn = Fnn
(c) Fpp > Fpn > Fnn
(d) Fpp < Fpn = Fnn

Answer:

(b) Fpp = Fpn = Fnn

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.

Fpp = Fpn = Fnn
Here, Fpp, Fpn and Fnn 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 Fn and Fe be the nuclear force and the electromagnetic force between them.
(a) Fe = Fn
(b) Fe >> Fn
(c) Fe << Fn
(d) Fe and Fn differ only slightly.

Answer:

(b) Fe >> Fn

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. Fe >> Fn

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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 Q-value 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 average-life,
(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 life-time 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 beta-decay (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 anti-neutrino.

Question 14:

A freshly prepared radioactive source of half-life 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 half-lives (as 26 = 64) of the radioactive source. Since the half-life 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 half-life and the average-life 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 half-life 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 N0.

No2=Noe-λt12⇒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 α-particle is bombarded on 14N. As a result, a 17O 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 (N-14) nucleus, an oxygen (O-17) 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 57Co kept in an open container beta-decays with a half-life 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

57Co 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 238U 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) xvt
(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 uranium-235 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, Ro = 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

≃ AmN
Here, mN 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 short-range nuclear forces on each other as well as on the neighbouring protons, whereas the protons exert attractive short-range 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 Q-value 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 Q-value of its decay reaction is negative. Hence, it is physically not possible.

Question 4:

Consider a sample of a pure beta-active 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) α-decay
(b) β+-decay
(c) β-decay
(d) γ-decay

Answer:

(d) γ-decay

​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→YZ-2A-4+H24e
During  β-decay​,  a neutron is converted to a proton​, 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+decay​ a proton in the nucleus is converted to a neutron​, 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→YZ-1A+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) α-decay
(b) β+-decay
(c) β-decay
(d) γ-decay

Answer:

(a) α-decay
(b) β+-decay

​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→YZ-2A-4+H24e
During  β-decay​,  a neutron is converted to a proton​, an electron and an antineutrino. Thus, there is an increase in the  atomic number.

XZA→YZ+1A+e-+ν¯
During  β+-decay​,  a proton in the nucleus is converted to a neutron​, a positron and a neutrino in order to maintain the stability of the nucleus. Thus, there is a decrease in the atomic number. ​

XZA→YZ-1A+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) α-rays
(b) beta-plus rays
(c) beta-minus 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, is the charge on the particle that is moving with speed v in a uniform magnetic field B.
Since alpha, beta-plus and beta-minus 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) α-rays
(b) Beta-plus rays
(c) Beta-minus rays
(d) Gamma rays

Answer:

(d) Gamma rays

Alpha rays, beta-plus and beta-minus 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 short-range 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 = Amp 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 = Amp
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

×1014 kg/m3

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 × 1030 kg (twice the mass of the sun).

Answer:

Given:
Mass of the neutron star, M = 4.0 × 1030 kg
Density of nucleus, d = 2.4

×1017

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 α-particle. 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+Nmn-Mc2Here, mp = Mass of proton
mn = 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:
7Li + p → α + α.
Atomic mass of 7Li = 7.0160 u and that of 4He = 4.0026 u.

Answer:

Given:
Mass of 7Li = 7.0160 u
Mass of 4He = 4.0026 u.
Reaction:

Li7+p→α+α+E,Energy release

Eis given by

E=mLi7+mp-2×mHe4c2=7.0160 u+1.007276 u-24.0026 uc2=(8.023273 u-8.0052 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+Nmn-M)c2
Here, mp = Mass of proton
M = Mass of nucleus
mn = Mass of neutron
c = Speed of light
On substituting the respective values, we get

B=[(79×1.007276+118×1.008665) u-196.96 u]c2  =198.597274-196.96×931 MeV  =1524.302094 MeVBinding energy per nucleon

=1524.3197=7.737 MeV

Question 6:

(a) Calculate the energy released if 238U emits an α-particle. (b) Calculate the energy to be supplied to 238U it two protons and two neutrons are to be emitted one by one. The atomic masses of 238U, 234Th and 4He are 238.0508 u, 234.04363 u and 4.00260 u respectively.

Answer:

(a)
Given:
Atomic mass of 238U, m(238U) = 238.0508 u
Atomic mass of 234Th, m(234Th)  = 234.04363 u
Atomic mass of 4He, m(4He) = 4.00260 u
When 238U emits an α-particle, the reaction is given by

U238→Th234+He4Mass defect, Δm = [m(238U)

-(m(234Th)+m(4He))]
Δm = [238.0508

-(234.04363 + 4.00260) = 0.00457 u
Energy released (E) when 238U emits an α-particle 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 238U 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
223Ra → 209Pb + 14C.
The atomic masses needed are as follows.
223Ra         209Pb        14C
22..018 u  208.981 u  14.003 u

Answer:

Given:
Atomic mass of 223Ra, m(223Ra) = 223.018 u
Atomic mass of 209Pb, m(209Pb) = 208.981 u
Atomic mass of 14C, m(14C) = 14.003 u

Reaction:

Ra223→209Pb+C14Energy, E=mR223a-mP209b+mC14c2                  = 223.018 u-208.981+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=(MZ-1,N+MB-MZ,N)c2where MZ,N = mass of an atom with Z protons and N neutrons in the nucleus and MB = mass of a hydrogen atom. This energy is known as proton-separation energy.

Answer:

Given:
Mass of an atom with Z protons and N neutrons = MZ,N
Mass of hydrogen atom = MH
As hydrogen contains only protons, the reaction will be given by

EZ,N→EZ-1,N+p1⇒EZ,N→Ez-1,N+1H1∴ Minimum energy needed to separate a proton,

∆E=(MZ-1,N+MH-MZ,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 MZ.N‘ MZ,N−1 and the mass of the neutron.

Answer:

Before the separation of neutron, let the mass of the nucleus be MZ,N.
Then, after the separation of neutron, the mass of the nucleus will be MZ,N-1.
The reaction is given by

EZ,N=EZ,N-1+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)c2

∆E=(MZ,N-1+MN-MZ,N)c2

Question 10:

32P beta-decays to 32S. Find the sum of the energy of the antineutrino and the kinetic energy of the β-particle. Neglect the recoil of the daughter nucleus. Atomic mass of 32P = 31.974 u and that of 32S = 31.972 u.

Answer:

Given:
Atomic mass of 32P, m(32P) = 31.974 u
Atomic mass of 32S, m(32S) = 31.972 u
Reaction:

P32→S32+1v0+-1β0Energy of antineutrino and β-particle, E = [m(32P)

m(32S)]c2
= (31.974 u− 31.972 u)c2
= 0.002 × 931 = 1.862 MeV

Question 11:

A free neutron beta-decays to a proton with a half-life of 14 minutes. (a) What is the decay constant? (b) Find the energy liberated in the process.

Answer:

Given:
Half-life period of free neutron beta-decays to a proton,

T1/2= 14 minutes

Half-life period, T1/2  =

0.6931λHere,

λ= Decay constant

∴ λ=0.69314×60      =8.25×10-4 s-1If mp is the mass of proton, let mn and me be the mass of neutron and mass of electron, respectively.

 

∴ Energy liberated, E=[mn-mp+me]c2                                       =[1.008665 u-1.007276+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 64Cu → 64Ni + 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=500×1.6×10-193×108×103         =2.67×10-22 Kgms-1

Question 14:

Potassium-40 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 Q-values 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 potassium-40 by βemission is given by
K4019→20Ca40+β-+v¯    Decay of potassium-40 by β+ emission is given by
K4019→18Ar40+β++v    Decay of potassium-40 by electron capture is given by
K4019+e-→18Ar40+v(b)
Qvalue in the β decay is given by
Qvalue = [m(19K40) − m(20Ca40)]c2
= [39.9640 u − 39.9626 u]c2
= 0.0014

×931 MeV
= 1.3034 MeV
Qvalue in the β+ decay is given by
Qvalue = [m(19K40) − m(20Ar40) − 2me]c2
= [39.9640 u − 39.9624 u −  0.0021944 u]c2
= (39.9640 − 39.9624) 931 MeV − 1022 keV
= 1489.96 keV − 1022 keV
= 0.4679 MeV
Qvalue in the electron capture is given by
Qvalue = [ m(19K40) − m(20Ar40)]c2
= (39.9640 − 39.9624)uc2
= 1.4890 = 1.49 MeV

Question 15:

Lithium (Z = 3) has two stable isotopes 6Li and 7Li. When neutrons are bombarded on lithium sample, electrons and α-particles 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¯+e-Be48→He24+He24

Question 16:

The masses of 11C and 11B are respectively 11.0114 u and 11.0093 u. Find the maximum energy a positron can have in the β*-decay of 11C to 11B.

Answer:

Given:
Mass of 11C, m(11C) = 11.0114 u
Mass of 11B, m(11B) = 11.0093 u
Energy liberated in the β+ decay (Q) is given by

Q=mC11-mB11-2mec2   = (11.0114 u − 11.0093 u

-2

×0.0005486 u)c2
= 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:

228Th emits an alpha particle to reduce to 224Ra. Calculate the kinetic energy of the alpha particle emitted in the following decay:

Th228→Ra224*+αRa224*→224Ra+Υ(217 keV).Atomic mass of 228Th is 228.028726 u, that of 224Ra is 224.020196 u and that of

H24is 4.00260 u.

Answer:

Given:
Atomic mass of 228Th, m(228Th) = 228.028726 u
Atomic mass of 224Ra, m(224Ra) = 224.020196 u
Atomic mass of

H24, m(

H24) = 4.00260 u
Mass of 224Ra = 224.020196 × 931 + 0.217 MeV = 208563.0195 MeV

Kinetic energy of alpha particle, K =

mTh228-mRa224+mH24c2
= (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:
12N → 12C* + e+ + v
12C* → 12C + γ (4.43MeV).
The atomic mass of 12N is 12.018613 u.

Answer:

Given:
Atomic mass of 12N, m(12N) = 12.018613 u
12N → 12C* + e+ + v
12C* → 12C + γ (4.43 MeV)

Net reaction is given by
 12N → 12C + e+ + v + γ (4.43 MeV)

Qvalue  of the

β+decay will be
Qvalue= [ m(12N)

-(m(12C*) + 2me)]c2
= [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 half-life? (b) What is the average-life? (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)
Half-life,

T12=0.693λ

⇒T1/2=0.6931.8×10-4            = 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
= N0
= 100
Active nuclei of mercury left after conversion of 25% isotope of mercury into gold = N =  75

Now,

NN0=e-λtHere, N = Number of inactive nuclei
N0 = Number of nuclei at t = 0
λ= Disintegration constant

On substituting the values, we get

75100=e-λt

⇒0.75=e-λx⇒ In 0.75=-λt⇒ t=In 0.75-0.00018       =1600 s

Page No 443:

Question 20:

The half-life of 199Au is 2.7 days. (a) Find the activity of a sample containing 1.00 µg of 198Au. (b) What will be the activity after 7 days? Take the atomic weight of 198Au to be 198 g mol−1.

Answer:

Given:
Half-life of 199Au, T1/2= 2.7 days

Disintegration constant,

λ=0.693T1/2=

0.6932.7×24×60×60=2.97×10-6  s-1Number of atoms left undecayed, N =

1×10-6×6.023×1023198Now, activity,

A0=λN
=

1×10-6×6.023×1023198×2.97×10-6   = 0.244 Ci

(b) After 7 days,
Activity,

A=A0e-λtHere, A0 = 0.244 Ci

∴ A =0.244×e-2.97×10-6×7×3600×24        =0.244×e17962.56×10-4       =0.040 Ci

Question 21:

Radioactive 131I has a half-life of 8.0 days. A sample containing 131I 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:
Half-life of radioactive 131I, T1/2 = 8 days
Activity of the sample at t = 0, A0 = 20 µCi
Time, t = 4 days

(a) Disintegration constant,

λ=0.693T1/2=0.0866Activity (A) at t = 4 days is given by

A=A0e-λt⇒A=20×10-6×e-0.0866×4      =1.41×10-6 Ci=14 μCi(b) Decay constant is a constant for a radioactive sample and depends only on its half-life.

λ=0.693[8×24×3600]    =1.0026×10-6 s-1

Question 22:

The decay constant of 238U is 4.9 × 10−18 S−1. (a) What is the average-life of 238U? (b) What is the half-life of 238U? (c) By what factor does the activity of a 238U sample decrease in 9 × 109 years?

Answer:

Given:
Decay constant,

λ= 4.9 × 10−18 s−1

(a) Average life of uranium

τis given by

τ=1λ  =14.9×10-18  =14.9×1018 s  =10164.9×365×24×36 years =10164.9×365×24×36  years = 6.47×10-7×1016 years =6.47×109 years(b) Half-life of uranium

T12is given by

T12=0.693λ=0.6934.9×10-18    =0.6934.9×1018 s    =0.1414×1018 s    =0.1414×1018365×24×3600    =1414×1012365×24×36   = 4.48×10-3×1012   =4.5×109 years(c) Time, t = 9 × 109 years
Activity (A) of the sample, at any time t, is given by
A=A02tT1/2  Here, A0 = 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 half-life?

Answer:

Given:
Initial rate of decay, A0 = 500,
Rate of decay after 50 minutes, A = 200
Time, t = 50 min
= 50

×60
= 3000 s

(a)
Activity, A = A0eλt
Here,

λ= Disintegration constant

∴ 200 = 500 × e−50×60×λ

⇒25=e-3000λ⇒ In25=-3000λ⇒ λ=3.05×10-4 s-1(b)

Half-life, T1/2=0.693λ                         =2272.13 s                         =38 min

Question 24:

The count rate from a radioactive sample falls from 4.0 × 106 per second to 1.0 × 106 per second in 20 hours. What will be the count rate 100 hours after the beginning?

Answer:

Given:
Initial count rate of radioactive sample, A0 = 4 × 106 disintegration/sec
Count rate of radioactive sample after 20 hours, A = 1 × 106 disintegration/sec
Time, t = 20 hours
Activity of radioactive sample

Ais given by

A=A02tT2Here, T1/2=Half-life 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 half-life of 226Ra is 1602 y. Calculate the activity of 0.1 g of RaCl2 in which all the radium is in the form of 226Ra. Taken atomic weight of Ra to be 226 g mol−1 and that of Cl to be 35.5 g mol−1.

Answer:

Given:
Half-life of radium, T1/2 = 1602 years
Atomic weight of radium = 226 g/mole
Atomic weight of chlorine = 35.5 g/mole
Now,
1 mole of RaCl2 = 226 + 71 = 297 g
297 g = 1 mole of RaCl2
0.1 g =

1297×0.1 mole of RaCl2Total number of atoms in 0.1 g of RaCl2, N

=0.1×6.023×1023297 = 0.02027×1022
∴ No of atoms, N = 0.02027

×1022

 

Disintegration constant, λ=0.693T12                                             =0.6931602×365×24×3600                                             =1.371×10-11Activity of radioactive sample, A =

λN                                                      =

1.371×10-11×2.027×1020                                                      =

2.8×109 disintegrations/second

Question 26:

The half-life 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:
Half-life of radioisotope,

T1/2= 10 hrs
Initial activity, A0 = 1 Ci
Disintegration constant,

λ=0.69310×3600 s-1Activity of radioactive sample,

A=A0e-λt
Here, A0 = Initial activity

λ= Disintegration constant
t = Time taken

After 9 hours,
Activity,

A=A0e-λt=1×e-0.69310×3600×9=0.536 Ci∴ Number of atoms left, N =

Aλ=

0.536×10×3.7×1010×36000.693=103.023×1013After 10 hrs,

Activity, A”=A0e-λt    =1×e-0.69310×10=0.5 CiNumber of atoms left after the 10th 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) × 1013
= 6.92 × 1013

Question 27:

The selling rate of a radioactive isotope is decided by its activity. What will be the second-hand rate of a one month old 32P(t1/2 = 14.3 days) source if it was originally purchased for 800 rupees?

Answer:

Given:
Half-life of 32P 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, A0 = 800 disintegration/sec
Disintegration constant

λis given by

λ=0.693T12=

0.69314.3 days-1Activity

Ais given by
A = A0eλt
Here,

λ= Disintegration constant

∴Activity of the radioactive isotope after one month (selling rate of the radioactive isotope)

Ais given below.

A=800×e-0.69314.3×30       =800×0.233669      =186.935=Rs 187

Question 28:

57Co decays to 57Fe by β+– emission. The resulting 57Fe is in its excited state and comes to the ground state by emitting γ-rays. The half-life of β+– decay is 270 days and that of the γ-emissions is 10−8 s. A sample of 57Co gives 5.0 × 109 gamma rays per second. How much time will elapse before the emission rate of gamma rays drops to 2.5 × 109 per second?

Answer:

According to the question, when the β+ decays to half of its original amount, the emission rate of γ-rays 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 βdecay? (b) The half-life of the decay scheme is 20.3 minutes. How much time will elapse before a mixture of 90% carbon-11 and 10% boron-11 (by the number of atoms) converts itself into a mixture of 10% carbon-11 and 90% boron-11?

Answer:

(a) The reaction is given by
C6→B5+β++v It is a β+ decay since atomic number is reduced by 1.

(b) Half-life of the decay scheme, T 1/2   = 20.3 minutes

Disintegration constant,

λ=0.693T12=

0.69320.3 min-1If t is the time taken by the mixture in converting, let the total no. of atoms be 100N0.

Carbon Boron
Initial 90 N0 10 N0
Final 10 N0 90 N0

N = N0eλt
Here, N0 = Initial number of atoms
N = Number of atoms left undecayed
10N0 = 90N0eλt  ( For carbon)

⇒ 19=e-0.69320.3×t  ⇒ In19=-0.69320.3t⇒ t= 64.36=64 min

Question 30:

4 × 1023 tritium atoms are contained in a vessel. The half-life 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, N0 = 4 × 1023
Half-life of tritium nuclei,

T12= 12.3 years

Disintegration constant,

λ=0.693T12=0.69312.3years

-1

Activity of the sample

Ais given by
A0 =

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

×1014 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=N0e-λt   Here, N0 = Initial number of atoms
∴ N=4×1023×e-0.69312.3×6.15=2.83×1023
Number of atoms disintegrated = (N0

N) = (4

-2.83) × 1023 = 1.17 × 1023

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 cm2 window. If the source contains 6.0 × 1016 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 × 1016
Total counts radiated from the source,

dNdt= Total surface area × 50000 counts/cm2
= 4 × 3.14 × 1 × 104 × 5 × 104
= 6.28 × 109 Counts
We know

dNdt=λNHere, λ = Disintegration constant

∴  λ=6.28×1096×1016       =1.0467×10-7       =1.05×10-7s-1

Question 32:

238U decays to 206Pb with a half-life of 4.47 × 109 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 238U and 0.600 mg of 206Pb. Assuming that all the lead has come from uranium, find the life of the rock.

Answer:

Given:
Half-life of 238U, t1/2 = 4.47 × 109 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 = N0e

-λt
Here,

λ= Disintegration constant
t = Life of the rock

⇒ N=N0e-0.693t1/2×t⇒ 0.0505=0.0680e-0.6934.47×109×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 14C per gram per minute. A sample from an ancient piece of charcoal shows 14C activity to be 12.3 disintegrations per gram per minute. How old is this sample? Half-life of 14C is 5730 y.

Answer:

Given:
Initial activity of charcoal, A0 = 15.3 disintegrations per gram per minute
Half-life 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  y-1Let 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=A0e-λt

A=A0e-0.6935730×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 beta-decays with a half-life 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:
Half-life time of tritium,

T12= 12.5 years
Disintegration constant,

λ=0.69312.5  per yearLet A0 be the activity, when the bottle was manufactured.
Activity after 8 years

Ais given by

A=A0e-0.69312.5×8     …(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’=A0e-λtHere, A’ = (Activity of the bottle manufactured 8 years ago) × 1.5%

A’=A0e-0.69312.5×8×0.015     …(2)Comparing (1) and (2)

0.69312.5 t=-0.6931×812.5+In [0.015]⇒ -0.69312.5 t=-0.69312.5×8-4.1997⇒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 128I varies with time as follows.

Time t (minute): 0 25 50 75 100
Ctount rate R (109 s−1): 30 16 8.0 3.8 2.0

(a) Plot In (R0/R) against t. (b) From the slope of the best straight line through the points, find the decay constant λ. (c) Calculate the half-life t1/2.

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 half-life period

T12is given by
T12=0.693λ     =0.6930.028=25 min

Question 36:

The half-life of 40K is 1.30 × 109 y. A sample of 1.00 g of pure KCI gives 160 counts s−1. Calculate the relative abundance of 40K (fraction of 40K present) in natural potassium.

Answer:

Given:
Half-life period of 40K,

T12= 1.30 × 109 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 × 1023 atoms are present in 40 gm.

Thus, 9.5 × 1018 atoms will be present in40×9.5×10186.023×1023 gm.=4×9.5×10-46.023 gm=6.309×10-4=0.00063 gmRelative abundance of 40K  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(Zb) with a = 4.95 × 107 s−1/2 and b = 1 to find the wavelength of the Kα X-ray 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(Zb)

We  know

v=cλHere, c = Speed of light
λ= Wavelength of the Kα X-ray

cλ=4.95×107(79-1)         =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 half-life t1/2. Show that after a time t >> t1/2 the number of active nuclei will become constant. Find the value of this constant.

Answer:

Given:
Half life period of isotope = t1/2

Disintegration constant,

λ=0.693t1/2Rate of Radio active decay

Ris given by,

R=dNdtWe are to show that after time t >> t1/2 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 N0.
Let N be the number of radio-active isotopes present at time t.
Then,
N = N0e−λt
Here,

λ= Disintegration constant
∴ Number of radioactive isotopes decayed = N0N = N0N0eλt
= N0 (1−eλt)   …(1)
Rate of decay

Ris given by
 R = λN0   …(2)
Substituting the value of N0 from equation (2) to equation (1), we get

N=N0(1-e-λt)  =Rλ(1-e-λt)

Question 40:

In an agricultural experiment, a solution containing 1 mole of a radioactive material (t1/2 = 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, N0 = 1 mole = 6 × 1023 atoms
Half-life of the radioactive material, T1/2 = 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 = N0e−λt
=6×1023×e-0.693×7014.3×24=6×1023×0.868=5.209×1023

Activity, R=dNdt=5.209×1023×0.69314.3×24  =0.0105×10233600dis/hr =2.9×10-6×1023 dis/sec =2.9×1017dis/secFraction of activity transmitted = 1 μCi2.9×1017×100%                                                     =1×3.7×1042.9×1017×100%                                                     =1.275×10-11%

Question 41:

A vessel of volume 125 cm3 contains tritium (3H, t1/2 = 12.3 y) at 500 kPa and 300 K. Calculate the activity of the gas.

Answer:

Given:
Volume of the vessel, V = 125 cm3 = 0.125 L
Half-life time of tritium, t1/2 = 12.3 y = 3.82 × 108 s
Pressure, P = 500 kpa = 5 atm
Temperature, T = 300 K,

Disintegration constant,

λ=0.693t1/2                                      =

0.6933.82×108=0.1814×10-8=1.81×10-9 s-1No. of atoms left undecayed, N = n × 6.023 × 1023
=5×0.1258.2×10-2×3×102×6.023×1023   ∵ n=PVRT=1.5×1022 atomsActivity, A = λN
= 1.81 × 10−9 × 1.5 × 1022 = 2.7 × 1013 disintegration/sec
∴A=2.7×10133.7×1010 Ci=729.72 Ci

Page No 444:

Question 42:

Bi83212can disintegrate either by emitting an α-particle of by emitting a β-particle. (a) Write the two equations showing the products of the decays. (b) The probabilities of disintegration α-and β-decays are in the ratio 7/13. The overall half-life of 212Bi is one hour. If 1 g of pure 212Bi is taken at 12.00 noon, what will be the composition of this sample at 1 P.m. the same day?

Answer:

Given:
Half-life of 212Bi, T1/2 = 1 h

-1
When

Bi83212disintegrates by emitting an α-particle

Bi83212→T812081+He24(α)When

Bi83212disintegrates by emitting a βparticle

Bi83212→P084212+β-+v¯Half-life period of 212Bi,

T12= 1 h

-1
At t = 0, the amount of 212Bi present = 1 g

At t = 1 = One half-life,
Amount of  212Bi present = 0.5 g
Probability of disintegration of α-decay and β-decay are in the ratio

713.
In 20 g of 212Bi, the amount of 208Ti formed = 7 g
In 1 g of 212Bi, the amount of 208Ti formed = 7/20 g
∴ Amount of 208Ti present in 0.5 g =

720×0.5=0.175 gIn 20 g of 212Bi, the amount of 212Po formed = 13 g
In 1 g of 212Bi, the amount of 212Po formed = 13/20 g
∴ Amount of 212Po present in 0.5 g =

1320×0.5=0.325 g

Question 43:

A sample contains a mixture of 108Ag and 110Ag isotopes each having an activity of 8.0 × 108 disintegration per second. 110Ag is known to have larger half-life than 108Ag. The activity A is measured as a function of time and the following data are obtained.

Time (s)

 

Activity (A)
(108 disinte-
grations s−1)
Time (s)

 

Activity (A
108 disinte-grations 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/A0) versus time. (b) See that for large values of time, the plot is nearly linear. Deduce the half-life of 110Ag from this portion of the plot. (c) Use the half-life of 110Ag to calculate the activity corresponding to 108Ag in the first 50 s. (d) Plot In (A/A0) versus time for 108Ag for the first 50 s. (e) Find the half-life of 108Ag.

Answer:

(a) Activity, A0 = 8 × 108 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) Half-life of 110Ag = 24.4 s
(c) Half-life of 110Ag,

T12= 24.4 s

Decay constant,

λ=0.693T12

λ=0.69324.4=0.0284∴ t = 50 sec

 

Activity, A=A0e-λt                 =8×108×e-0.0284×50                   =1.93×108(d)

(e) The half-life period of 108Ag 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 99Tc. This isotope is radioactive with a half-life 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, t1 = 24 hours
Half-life of radioactive isotope, t2 = 6 hours
Initial activity, A0 = 6 μCi
Let after time t, activity of the sample be A.
Half-life 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 average-life 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=qe-t/CRHere, Q = Charge left
q = Initial charge
C = Capacitance
R = Resistance

Energy, E =

12Q2C=

q2e-2t/CR2CActivity, A = A0e

-λt
Here, A0 = Initial activity

λ= Disintegration constant

∴ Ratio of the energy to the activity =

EA=q2×e-2t/CR2CA0e-λtSince 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 half-life 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=i01-e-RtLNumber of active nuclei

Nat any time

tis given by

N=N0e-λtWhere N0 = Total number of nuclei

λ= Disintegration constant
Now,

iN=i01-e-tR/LN0e-λtAs

iNis independent of time, coefficients of t are equal.

Let

t12be the half-life of the isotope.

 

⇒-RL=-λ⇒RL=0.693t12⇒t12=0.693×10-3=6.93×10-4 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 235U has an abundance of 0.7% by weight in natural uranium.

Answer:

235 g of uranium contains 6.02 × 1023 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 =

6.023×1023×0.007×200×106×1.6×10-19235 J= 5.74

×108 J

Question 48:

A uranium reactor develops thermal energy at a rate of 300 MW. Calculate the amount of 235U 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

×106

×1.6

×10

-19= Power
N

×200

×106

×1.6

×10

-19= 300

×106

⇒N=32×1.6×1019 = 33.2×1019  atoms
6.023 × 1023 atoms = 238 gm of U235

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 235U, at what rate will the amount of 235U decrease? Express your answer in kg per day. (c) Assuming that uranium enriched to 3% in 235U will be used, how much uranium is needed per month (30 days)?

Answer:

(a) Total population of the town = 1 million =  106
Average electric power needed per person = 300 W
Total power used by the town in one day = 300 × 106 × 60 × 60 × 24 J = 300 × 86400 ×106 J
Energy generated in one fission = 200 × 106 × 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=3.2×10-11×25100E=8×10-12 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 × 106 J
N = 3.24 × 1024

(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 235U be used.

⇒ x×3100 = 1.264×30⇒x = 1264 kg

Question 50:

Calculate the Q-values 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 H12-mH33+mH13c2         =(4.028204-4.023874)×931 MeV         =4.05 MeV(b) Q=2×m H12-(m H23+mn)c2         =[4.028204-4.024694)×931         =0.00351×931         =3.25 MeV(c) Q=m H12+m H13-m He14-mnc2        =(2.014102+3.016049-4.002603-1.008665)×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

×109

Charge,

q1=q2=2×1.6×10-19 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     =

2×9×109×4×1.6×10-1923×1.38×10-23×2×10-15     = 2.23

×1010 K

Question 52:

Calculate the Q-value of the fusion reaction
4He + 4He = 8Be.
Is such a fusion energetically favourable? Atomic mass of 8Be is 8.0053 u and that of 4He is 4.0026 u.

Answer:

Given:
Atomic mass of 8Be = 8.0053 u
Atomic mass of 4He = 4.0026 u

Required Q-value=(2×4.0026-8.0053) 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
2H + 2H → 3H + p.
Assume that 1.5 × 10−2% of natural water is heavy water D2O (by number of molecules) and all the deuterium is used for fusion.

Answer:

Given:

18 g of water contains 6.023

×1023 molecules.

∴ 1000 g of water =

6.023×1023×100018=3.346×1025molecules
% of deuterium =

3.346×1025 × 0.015100= 0.05019

×1023
Energy of deuterium =

30.4486×1025

=2×mH2-mH3-mpc2=2×2.014102 u-3.016049 u-1.007276 uc2=0.004879×931 MeV=4.542349 MeV=7.262 ×10-13 J    Total energy = 0.05019

×1023

×7.262

×10

-13 J
= 3644 MJ

HC Verma Solutions for Class 11 Physics – Part 1

HC Verma Solutions for Class 12 Physics – Part 2