
Chapter 43 – Bohr’s Model and Physics of the Atom
Page No 383:
Question 6:
In which of the following systems will the wavelength corresponding to n = 2 to n = 1 be minimum?
(a) Hydrogen atom
(b) Deuterium atom
(c) Singly ionized helium
(d) Doubly ionized lithium
Answer:
(d) Doubly ionized lithium
The wavelength corresponding the transition from n_{2} to n_{1} is given by
1λ=RZ21n121n22Here,
R = Rydberg constant
Z = Atomic number of the ion
From the given formula, it can be observed that the wavelength is inversely proportional to the square of the atomic number.
Therefore, the wavelength corresponding to n = 2 to n = 1 will be minimum in doubly ionized lithium ion because for lithium, Z = 3.
Page No 382:
Question 1:
How many wavelengths are emitted by atomic hydrogen in visible range (380 nm − 780 nm)? In the range 50 nm to 100 nm?
Answer:
Balmer series contains wavelengths ranging from 364 nm (for n_{2} = 3) to 655 nm (n_{2} =
∞).
So, the given range of wavelength (380−780 nm) lies in the Balmer series.
The wavelength in the Balmer series can be found by
1λ=R1221n2Here, R = Rydberg’s constant = 1.097×10^{7} m
1
The wavelength for the transition from n = 3 to n = 2 is given by
1λ1=R122132λ1=656.3 nmThe wavelength for the transition from n = 4 to n = 2 is given by
1λ2=R122142λ2=486.1 nmThe wavelength for the transition from n = 5 to n = 2 is given by
1λ3=R122152λ3=434.0 nmThe wavelength for the transition from n = 6 to n = 2 is given by
1λ4=R122162λ4=410.2 nmThe wavelength for the transition from n = 7 to n = 2 is given by
1λ5=R122172λ5=397.0 nmThus, the wavelengths emitted by the atomic hydrogen in visible range (380−780 nm) are 5.
Lyman series contains wavelengths ranging from 91 nm (for n_{2} = 2) to 121 nm (n_{2 }=
∞).
So, the wavelengths in the given range (50−100 nm) must lie in the Lyman series.
The wavelength in the Lyman series can be found by
1λ=R1121n2The wavelength for the transition from n = 2 to n = 1 is given by
1λ1=R112122λ1=122 nmThe wavelength for the transition from n = 3 to n = 1 is given by
1λ2=R112122λ2=103 nmThe wavelength for the transition from n = 4 to n = 1 is given by
1λ3=R112142λ3=97.3 nmThe wavelength for the transition from n = 5 to n = 1 is given by
1λ4=R112152λ4=95.0 nmThe wavelength for the transition from n = 6 to n = 1 is given by
1λ5=R112162λ5=93.8 nmSo, it can be noted that the number of wavelengths lying between 50 nm to 100 nm are 3.
Question 2:
The first excited energy of a He^{+} ion is the same as the ground state energy of hydrogen. Is it always true that one of the energies of any hydrogenlike ion will be the same as the ground state energy of a hydrogen atom?
Answer:
The energy of hydrogen ion is given by
En=(13.6 eV) Z2n2For the first excited state (n = 2), the energy of He^{+} ion (with Z = 2) will be
13.6 eV. This is same as the ground state energy of a hydrogen atom.
Similarly, for all the hydrogen like ions, the energy of the (n
1)th excited state will be same as the ground state energy of a hydrogen atom if Z = n.
Question 3:
Which wavelengths will be emitted by a sample of atomic hydrogen gas (in ground state) if electrons of energy 12.2 eV collide with the atoms of the gas?
Answer:
As the electron collides, it transfers all its energy to the hydrogen atom.
The excitation energy to raise the electron from the ground state to the nth state is given by
E=(13.6 eV)×1121n2Substituting n = 2, we get
E = 10.2 eV
Substituting n = 3, we get
E‘ = 12.08 eV
Thus, the atom will be raised to the second excited energy level.
So, when it comes to the ground state, there is transitions from n = 3 to n = 1.
Therefore, the wavelengths emitted will lie in the Lyman series (infrared region).
Question 4:
When white radiation is passed through a sample of hydrogen gas at room temperature, absorption lines are observed in Lyman series only. Explain.
Answer:
White radiations are xrays that have energy ranging between 510 eV.
When white radiation is passed through a sample of hydrogen gas at room temperature, absorption lines are observed only in the Lyman series.
At room temperature, almost all the atoms are in ground state.
The minimum energy required for absorption is 10.2 eV (for a transition from n = 1 to n = 2).
The white radiation has photon radiations that have an energy of around 10.2 eV.
So, they are just sufficient to transmit an electron from n = 1 to n = 2 level.
Hence, the absorption lines are observed only in the Lyman series.
Page No 383:
Question 5:
Balmer series was observed and analysed before the other series. Can you suggest a reason for such an order?
Answer:
The Balmer series lies in the visible range. Therefore, it was observed and analysed before the other series. The wavelength range of Balmer series is from 364 nm (for n_{2} =
∞) to 655 nm (for n_{2 }=3).
Question 6:
What will be the energy corresponding to the first excited state of a hydrogen atom if the potential energy of the atom is taken to be 10 eV when the electron is widely separated from the proton? Can we still write E_{n} = E_{1}/n^{2}, or r_{n} = a_{0} n^{2}?
Answer:
Energy of n^{th} state of hydrogen is given by
En = 13.6n2 eVEnergy of first excited state (n = 2) of hydrogen, E_{1} =
13.64 eV= 3.4 eV
This relation holds true when the refrence point energy is zero.Usually the refrence point energy is the energy of the atom when the electron is widely separated from the proton.In the given question, the potential energy of the atom is taken to be 10 eV when the electron is widely separated from the proton so here our refrence point energy is 10 eV. Earlier The energy of first excited state was 3.4 eV when the refrence point had zero energy but now as the refrence point has shifted so The energy of the first excited state will also shift by the corresponding amount.Thus,
E_{1}^{,} = 3.4 eV10 eV = 13.4 eV
We still write E_{n} = E_{1}/n^{2}, or r_{n} = a_{0} n^{2} because these formulas are independent of the refrence point enegy.
Question 7:
The difference in the frequencies of series limit of Lyman series and Balmer series is equal to the frequency of the first line of the Lyman series. Explain.
Answer:
The ‘series limit’ refers to the ‘shortest wavelength’ (corresponding to the maximum photon energy).
The frequency of the radiation emitted for transition from n_{1} to n_{2}_{ }is given by
f=k1n121n22Here, k is a constant.
For the series limit of Lyman series,
n_{1} = 1
n_{2} =
∞Frequency,
f1=k1121∞=kFor the first line of Lyman series,
n_{1} = 1
n_{2} = 2
Frequency,
f2=k112122=3k4For series limit of Balmer series,
n_{1} = 2
n_{2} =
∞ f1=k1221∞=k4f_{1}
–f_{3} = f_{2}
Thus, the difference in the frequencies of series limit of Lyman series and Balmer series is equal to the frequency of the first line of the Lyman series.
Question 8:
The numerical value of ionization energy in eV equals the ionization potential in volts. Does the equality hold if these quantities are measured in some other units?
Answer:
The electron volt is the amount of energy given to an electron in order to move it through the electric potential difference of one volt.
1 eV = 1.6 × 10^{–19} J
The numerical value of ionisation energy in eV is equal to the ionisation potential in volts. The equality does not hold if these quantities are measured in some other units.
Question 9:
We have stimulated emission and spontaneous emission. Do we also have stimulated absorption and spontaneous absorption?
Answer:
When a photon of energy (E_{2} – E_{1} = hυ) is incident on an atom in the ground state, the atom in the ground state E_{1} may absorb the photon and jump to a higher energy state (E_{2}). This process is called stimulated absorption or induced absorption.
Spontaneous absorption is the process by which an atom in its ground state spontaneously jumps to a higher energy state, resulting in the absorption of a photon.
We do not have any process such as spontaneous absorption. This is because for absorption, we need to incident a photon of sufficient energy on the atom to stimulate the atom for absorption.
Question 10:
An atom is in its excited state. Does the probability of its coming to ground state depend on whether the radiation is already present or not? If yes, does it also depend on the wavelength of the radiation present?
Answer:
When an atom transits from an excited state to ground state in the presence of an external radiation then it is called as stimulated transition.
When an atom transits from an excited state to ground state on its own then it is called as spontaneous transition.
Ratio of the coefficient for stimulated transition to spontaneous transition is given by
R = ABρ(ν) = ehνkT1For microwave region
ν = 1010 (say)R = e6.6×1034×10101.38×1023×3001=e0.00161=0.0016This implies that stimulated transition dominate in this region.
For visible region
ν = 1015⇒R = e1601⇒R>>1So here spontaneous transition dominate.
Question 1:
The minimum orbital angular momentum of the electron in a hydrogen atom is
(a) h
(b) h/2
(c) h/2π
(d) h/λ
Answer:
(c) h/2π
According to Bohr’s atomic theory, the orbital angular momentum of an electron is an integral multiplt of h/2π.
∴
Ln=nh2πHere,
n = Principal quantum number
The minimum value of n is 1.
Thus, the minimum value of the orbital angular momentum of the electron in a hydrogen atom is given by
L=h2π
Question 2:
Three photons coming from excited atomichydrogen sample are picked up. Their energies are 12.1 eV, 10.2 eV and 1.9 eV. These photons must come from
(a) a single atom
(b) two atoms
(c) three atoms
(d) either two atoms or three atoms
Answer:
(d) either two atoms or three atoms
The energies of the photons emitted can be expressed as follows:
13.6112122 eV = 10.2 eV
13.6112132 eV = 12.1 eV
13.6122132 eV = 1.9 eVThe following table gives the transition corresponding to the energy of the photon:
Energy of photon  Transition 
12.1 eV  n = 3 to n = 1 
10.2 eV  n = 2 to n = 1 
1.9 eV  n = 3 to n = 2 
A hydrogen atom consists of only one electron. An electron can have transitions, like from n = 3 to n = 2 or from n = 2 to n = 1, at a time.
So, it can be concluded that the photons are emitted either from three atoms (when all the three transitions of electrons are in different atoms) or from two atoms (when an atom has n = 3 to n = 2 and then n = 2 to n = 1 electronic transition and the other has n = 3 to n = 1 electronic transition).
Question 3:
Suppose, the electron in a hydrogen atom makes transition from n = 3 to n = 2 in 10^{−8} s. The order of the torque acting on the electron in this period, using the relation between torque and angular momentum as discussed in the chapter on rotational mechanics is
(a) 10^{−34} N m
(b) 10^{−24} N m
(c) 10^{−42} N m
(d) 10^{−8} N m
Answer:
(c) 10^{−42} Nm
The angular momentum of the electron for the nth state is given by
Ln=nh2πAngular momentum of the electron for n = 3,
Li=3h2πAngular momentum of the electron for n = 2,
Lf=2h2πThe torque is the time rate of change of the angular momentum.
Torque, τ=LfLit =(2h/2π)(3h/2π)108 =(h/2π)108 =1034108 ∵h2π≈1034 Js =1042 NmThe magnitude of the torque is 10
42 Nm.
Question 4:
In which of the following transitions will the wavelength be minimum?
(a) n = 5 to n = 4
(b) n = 4 to n = 3
(c) n = 3 to n = 2
(d) n = 2 to n = 1
Answer:
(d) n = 2 to n = 1
For the transition in the hydrogenlike atom, the wavelength of the emitted radiation is calculated by
1λ=RZ21n11n2Here, R is the Rydberg constant.
For the transition from n = 5 to n = 4, the wavelength is given by
1λ=RZ2142152λ=4009RZ2For the transition from n = 4 to n = 3, the wavelength is given by
1λ=RZ2132142λ=1447RZ2For the transition from n = 3 to n = 2, the wavelength is given by
1λ=RZ2122132λ=365RZ2For the transition from n = 2 to n = 1, the wavelength is given by
1λ=RZ2112122λ=2RZ2From the above calculations, it can be observed that the wavelength of the radiation emitted for the transition from n = 2 to n = 1 will be minimum.
Question 5:
In which of the following systems will the radius of the first orbit (n = 1) be minimum?
(a) Hydrogen atom
(b) Deuterium atom
(c) Singly ionized helium
(d) Doubly ionized lithium
Answer:
(d) Doubly ionized lithium
For a hydrogenlike ion with Z protons in the nucleus, the radius of the nth state is given by
rn=n2a0ZHere, a_{0} = 0.53 pm
For lithium,
Z = 3
Therefore, the radius of the first orbit for doubly ionised lithium will be minimum.
Question 7:
Which of the following curves may represent the speed of the electron in a hydrogen atom as a function of the principal quantum number n?
Figure
Answer:
(c)
The speed (v) of electron can be expressed as
v=Ze22∈0hn ….(1)
Here,
Z = Number of protons in the nucleus
e = Magnitude of charge on electron charge
n = Principal quantum number
h = Planck’s constant
It can be observed from equation (1) that the velocity of electron is inversely proportional to the principal quantum number (n).
Therefore, the graph between them must be a rectangular hyperbola.
The correct curve is (c).
Question 8:
As one considers orbits with higher values of n in a hydrogen atom, the electric potential energy of the atom
(a) decreases
(b) increases
(c) remains the same
(d) does not increase
Answer:
(b) increases
The electric potential energy of hydrogen atom with electron at the nth state is given by
V=2×13.6n2As the value of n increases, the potential energy of the hydrogen atom also increases, i.e. the atom becomes less bound as n increases.
Question 9:
The energy of an atom (or ion) in its ground state is −54.4 eV. It may be
(a) hydrogen
(b) deuterium
(c) He^{+}
(d) Li^{++}
Answer:
(c) He^{+}
The total energy of a hydrogenlike ion, having Z protons in its nucleus, is given by
E=13.6Z2n2eV
Here, n = Principal quantum number
For ground state,
n = 1
∴ Total energy, E =
13.6 Z^{2} eV
For hydrogen,
Z = 1
∴ Total energy, E =
13.6 eV
For deuterium,
Z = 1
∴ Total energy, E =
13.6 eV
For He^{+},
Z = 2
∴ Total energy, E =
13.6×2^{2} =
54.4 eV
For Li^{++},
Z = 3
∴ Total energy, E =
13.6×3^{2} =
122.4 eV
Hence, the ion having an energy of −54.4 eV in its ground state may be He^{+}.
Question 10:
The radius of the shortest orbit in a oneelectron system is 18 pm. It may be
(a) hydrogen
(b) deuterium
(c) He^{+}
(d) Li^{++}
Answer:
(d) Li^{++}
The radius of the nth orbit in one electron system is given by
rn=n2a0ZHere, a_{0} = 53 pm
For the shortest orbit,
n = 1
For hydrogen,
Z = 1
∴ Radius of the first state of hydrogen atom = 53 pm
For deuterium,
Z= 1
∴ Radius of the first state of deuterium atom = 53 pm
For He^{+},
Z = 2
∴ Radius of He^{+} atom =
532 pm =26.5 pmFor Li^{++},
Z = 3
∴ Radius of Li^{++} atom =
533 pm =17.66 pm≈18 pmThe given oneelectron system having radius of the shortest orbit to be 18 pm may be Li^{++}.
Question 11:
A hydrogen atom in ground state absorbs 10.2 eV of energy. The orbital angular momentum of the electron is increased by
(a) 1.05 × 10^{−34} J s
(b) 2.11 × 10^{−34} J s
(c) 3.16 × 10^{−34} J s
(d) 4.22 × 10^{−34} J s
Answer:
(a) 1.05 × 10^{−34} J s
Let after absorption of energy, the hydrogen atom goes to the nth excited state.
Therefore, the energy absorbed can be written as
10.2=13.6×1121n2⇒10.213.6=11n2⇒1n2=13.610.213.6⇒1n2=3.413.6⇒n2=4⇒n=2The orbital angular momentum of the electron in the nth state is given by
Ln=nh2πChange in the angular momentum,
∆L=2h2πh2π=h2π∴
∆L=1.05×1034 Js
Question 12:
Which of the following parameters are the same for all hydrogenlike atoms and ions in their ground states?
(a) Radius of the orbit
(b) Speed of the electron
(c) Energy of the atom
(d) Orbital angular momentum of the electron
Answer:
(d) Orbital angular momentum of the electron
According to Bohr’s atomic theory, the orbital angular momentum of an electron in a oneelectron system is given by
Ln=nh2πHere,
n = Principal quantum number
The angular momentum is independent of the atomic number of the oneelectron system. Therefore, it is same for all hydrogenlike atoms and ions in their ground states.
The other parameters given here are dependent on the atomic number of the hydrogenlike atom or ion taken.
Question 13:
In a laser tube, all the photons
(a) have same wavelength
(b) have same energy
(c) move in same direction
(d) move with same speed
Answer:
(d) move with same speed
All the photons emitted in the laser move with the speed equal to the speed of light (c = 3×10^{8} m/s).
Ideally, the light wave through the laser must be coherent, but in practical laser tubes, there is some deviation from the ideal result. Thus, the photons emitted by the laser have little variations in their wavelengths and energies as well as the directions, but the velocity of all the photons remains same.
Question 1:
In a laboratory experiment on emission from atomic hydrogen in a discharge tube, only a small number of lines are observed whereas a large number of lines are present in the hydrogen spectrum of a star. This is because in a laboratory
(a) the amount of hydrogen taken is much smaller than that present in the star
(b) the temperature of hydrogen is much smaller than that of the star
(c) the pressure of hydrogen is much smaller than that of the star
(d) the gravitational pull is much smaller than that in the star
Answer:
(b) the temperature of hydrogen is much smaller than that of the star
The number of lines of the hydrogen spectrum depends on the excitation of the hydrogen atom. This is dependent on the heat energy absorbed by the hydrogen atoms. More the temperature of the hydrogen sample, more is the heat energy. The temperature of hydrogen at the star is much more than that can be produced in the laboratory. Hence, less number of lines are observed in the hydrogen spectrum in the laboratory than that in a star.
Page No 384:
Question 2:
An electron with kinetic energy 5 eV is incident on a hydrogen atom in its ground state. The collision
(a) must be elastic
(b) may be partially elastic
(c) must be completely inelastic
(d) may be completely inelastic
Answer:
(a) must be elastic.
The minimum energy required to excite a hydrogen atom from its ground state to 1st excited state is approximately 10 eV. As the incident electron energy is not sufficient for excitation of the hydrogen atom so electron will not get absorbed in the hydrogen atom so it can not be an inelastic collision. Also this collision can not be partially elastic because in an partially elestic collision, there is a net loss on kinetic energy. If the energy is lost then corresponding amount of heat shlould have been produced but it is not so which implies that the collision is completely elastic.
Question 3:
Which of the following products in a hydrogen atom are independent of the principal quantum number n? The symbols have their usual meanings.
(a) vn
(b) Er
(c) En
(d) vr
Answer:
(a) vn
(b) Er
Relations for energy, radius of the orbit and its velocity are given by
E = mZ2e48∈02h2n2r = ∈0h2n2πmZe2v= Ze22∈0hnWhere
Z : the atomic number of hydrogen like atom
e : electric charge
h : plank constant
m : mass of electron
n : principal quantam number of the electron
∈0: permittivity of vacuum
From these relations, we can see that the products independent of n are vn, Er.
Question 4:
Let A_{n} be the area enclosed by the nth orbit in a hydrogen atom. The graph of ln (A_{n}/A_{1}) against ln(n)
(a) will pass through the origin
(b) will be a straight line with slope 4
(c) will be a monotonically increasing nonlinear curve
(d) will be a circle
Answer:
(a) will pass through the origin
(b) will be a straight line with slope 4
The radius of the nth orbit of a hydrogen atom is given by
rn=n2a0Area of the nth orbit is given by
An = πrn2 = πn4a02A1 = πa02⇒lnAnA1= lnπn4a02πa02lnAnA1= 4ln n …(1)From the above expression, the graph of ln (A_{n}/A_{1}) against ln(n) will be a straight line passing through the origin and having slope 4.
Question 5:
Ionization energy of a hydrogenlike ion A is greater than that of another hydrogenlike ion B. Let r, u, E and L represent the radius of the orbit, speed of the electron, energy of the atom and orbital angular momentum of the electron respectively. In ground state
(a) r_{A} > r_{B}
(b) u_{A} > u_{B}
(c) E_{A} > E_{B}
(d) L_{A} > L_{B}
Answer:
(b) u_{A} > u_{B}
The ionisation energy of a hydrogen like ion of atomic number Z is given by
V=(13.6 eV)×Z2Thus, the atomic number of ion A is greater than that of B (Z_{A} > Z_{B}).
The radius of the orbit is inversely proportional to the atomic number of the ion.
∴ r_{A} > r_{B}
Thus, (a) is incorrect.
The speed of electron is directly proportional to the atomic number.
Therefore, the speed of the electron in the orbit of A will be more than that in B.
Thus, u_{A} > u_{B} is correct.
The total energy of the atom is given by
E=mZ2e28∈0h2n2As the energy is directly proportional to Z^{2}, the energy of A will be less than that of B, i.e. E_{A} < E_{B}.
The orbital angular momentum of the electron is independent of the atomic number.
Therefore, the relation L_{A} > L_{B }is invalid.
Question 6:
When a photon stimulates the emission of another photon, the two photons have
(a) same energy
(b) same direction
(c) same phase
(d) same wavelength
Answer:
(a) same energy
(b) same direction
(c) same phase
(d) same wavelength
When a photon stimulates the emission of another photon, the two photons have same energy, direction, phase, and wavelength or we can say that the two photons are coherent.
When an atom is present in its excited state then if a photon of energy equal to the energy gap between the excited state and any lower stable state is incident on this atom then the atom transits from upper state to the lower stable state by emitting a photon of energy equal to the energy gap between the two states. It is called stimulated emission. The emitted photon and incident photon have same energy and hence same wavelength. Also these two photons will be in phase and in the same direction. This process of producing monochromatic and unidirectional light is called lasing action.
Question 1:
The Bohr radius is given by
a0=ε0h2πme2. Verify that the RHS has dimensions of length.
Answer:
The dimensions of ε_{0} can be derived from the formula given below:
a=ε0 h2πme2=A2T2ML2T12L2ML2MAT2=M2L2T2M2L3T2=LClearly, a_{0} has the dimensions of length.
Question 2:
Find the wavelength of the radiation emitted by hydrogen in the transitions (a) n = 3 to n = 2, (b) n = 5 to n = 4 and (c) n = 10 to n = 9.
Answer:
From Balmer empirical formula, the wavelength
λof the radiation is given by
1λ=R1n121n22Here, R = Rydberg constant = 1.097
×10^{7 }m
1
n_{1} = Quantum number of final state
n_{2} = Quantum number of initial state
(a)
For transition from n = 3 to n = 2:
Here,
n_{1} = 2
n_{2} = 3
1λ=1.09737×107×1419⇒ λ=365×1.09737×107 =6.56×107=656 nm(b)
For transition from n = 5 to n = 4:
Here,
n_{1} = 4
n_{2} = 5
1λ=1.09737×107 116125⇒ λ=4001.09737×107×9 =4050 nm(c)
For transition from n = 10 to n = 9:
Here,
n_{1} = 9
n_{2} = 10
1λ=1.09737×107 1811100λ=81×10019×1.09737×107 =38849 nm
Question 3:
Calculate the smallest wavelength of radiation that may be emitted by (a) hydrogen, (b) He^{+} and (c) Li^{++}.
Answer:
Given:
For the smallest wavelength, energy should be maximum.
Thus, for maximum energy, transition should be from infinity to the ground state.
∴ n_{1} = 1
n_{2}=
∞
(a) Wavelength of the radiation emitted
λis given by
1λ=RZ21n121n22For hydrogen,
Atomic number, Z = 1
R = Rydberg constant = 1.097×10^{7} m
1
On substituting the respective values,
λ=11.097×107=11.097×107 =0.911×107 =91.1×109=91 nm(b)
For He^{+},
Atomic number, Z = 2
Wavelength of the radiation emitted by He^{+ }(
λ) is given by
1λ=RZ21n121n22
∴1λ=22(1.097×107) 1121∞2
⇒ λ=91 nm4=23 nm(c) For Li^{++},
Atomic number, Z = 3
Wavelength of the radiation emitted by Li^{++ }(
λ) is given by
1λ=RZ21n121n22
∴1λ=32×(1.097×107) 1121∞2
⇒λ=91 nmZ2=919=10 nm
Question 4:
Evaluate Rydberg constant by putting the values of the fundamental constants in its expression.
Answer:
Expression of Rydberg constant (R) is given by
R=me48h2c ∈02Mass of electron, m_{e} = 9.31
×10^{21} kg
Charge, e = 1.6 × 10^{−19} C
Planck’s constant, h = 6.63 × 10^{−34} Js,
Speed of light, c = 3 × 10^{8} m/s,
Permittivity of vacuum, ∈_{0} = 8.85 × 10^{−12}^{ }C^{2}N
1m
On substituting the values in the expression, we get
R=9.31×1031×1.6×101948×6.63×10342×3×108×8.85×10122⇒R =1.097×107 m1
Question 5:
Find the binding energy of a hydrogen atom in the state n = 2.
Answer:
The binding energy (E) of hydrogen atom is given by
E=13.6n2 eVFor state n = 2,
E = 13.622⇒E=3.4 eVThus, binding energy of hydrogen at n = 2 is
3.4 eV.
Question 6:
Find the radius and energy of a He^{+} ion in the states (a) n = 1, (b) n = 4 and (c) n = 10.
Answer:
For He^{+} ion,
Atomic number, Z = 2
For hydrogen like ions, radius
rof the nth state is given by
r=0.53 n2ZÅ
Here,
Z = Atomic number of ions
n = Quantum number of the state
Energy
Eof the nth state is given by
E_{n} =
13.6 Z2n2(a)
For n = 1,
Radius,
r=0.53×122 = 0.265 A0 Å
Energy, E_{n} =
13.6×41 =
54.4 eV
(b)
For n = 4,
Radius,
r=0.53×162=4.24 ÅEnergy,
E=13.6×416=3.4 eV(c)
For n = 10,
Radius,
r=0.53×1002 = 26.5 Å
Energy,
E=13.6×4100=0.544 eV
Question 7:
A hydrogen atom emits ultraviolet radiation of wavelength 102.5 nm. What are the quantum numbers of the states involved in the transition?
Answer:
Wavelength of ultraviolet radiation,
λ= 102.5 nm = 102,5
×10
9m
Rydberg’s constant, R = 1.097
×10^{7} m
1
Since the emitted light lies in ultraviolet range, the lines will lie in lyman series.
Lyman series is obtained when an electron jumps to the ground state (n_{â€‹1 }= 1) from any excited state (n_{â€‹2}).
Wavelength of light
λis given by
1λ=R1n121n22Here, R=Rydberg constant⇒1102.5×109=1.097×107 1121n22⇒109102.5=1.097×107 11n22⇒102102.5=1.097 11n22⇒11n22=100102.5×1.097⇒1n22=1100102.5×1.097⇒n2=9.041=3The transition will be from 1 to 3.
Question 8:
(a) Find the first excitation potential of He^{+} ion. (b) Find the ionization potential of Li^{++} ion.
Answer:
(a) PE of hydrogen like atom in the nth state, V =
=13.6Z2n2eVHere, Z is the atomic number of that atom.
For the first excitation, the atom has to be excited from n = 1 to n = 2 state.
So, its excitation potential will be equal to the difference in the potential of the atom in n = 1 and in n = 2 states.
First excitation potential of He^{+}
13.6Z2 (1122) eV=10.2 Z2 eV⇒10.2 × Z^{2}
= 10.2 × 4
= 40.8 V
(b) Ionization Potential Li^{++} = 13.6 V × Z^{2}
= 13.6 × 9
= 122.4 V
Question 9:
A group of hydrogen atoms are prepared in n = 4 states. List the wavelength that are emitted as the atoms make transitions and return to n = 2 states.
Answer:
There will be three wavelengths.
(i) For the transition from (n = 4) to (n = 3) state
(ii) For the transition from (n = 3) to (n = 2) state
(iii) For the transition from (n = 4) to (n = 2) state
Let (
λ_{1}) be the wavelength when the atom makes transition from (n = 4) state to (n = 2) state._{â€‹ }
Here,
n_{1}_{ }= 2
n_{2}_{ }= 4
Now, the wavelength (
λ_{1})â€‹ will be
1λ1=R1n121n22
R=1.097×107 m11λ1=1.097×107×14116⇒1λ1=1.097×107 4116⇒1λ1=1.097×107×316⇒λ1=16×1073×1.097 = 4.8617×107 = 486.1×109 =487 nm
When an atom makes transition from (n = 4) to (n = 3), the wavelength (
λ2) is given by
Here again, n1=3 n2=41λ2=1.097×107 19116⇒1λ2=1.097×107 169144⇒1λ2=1.097×107×7144⇒λ2=1441.097×107×7 =1875 nmSimilarly, wavelength (
λ_{3}) for the transition from (n = 3) to (n = 2) is given by
When the transition is n_{1} = 2 to n_{2} = 3:
1λ3=1.097×1071419⇒1λ3=1.097×107 9436⇒1λ3=1.097×107×536⇒λ3=36×1071.097×5=656 nm
Question 10:
A positive ion having just one electron ejects it if a photon of wavelength 228 Å or less is absorbed by it. Identify the ion.
Answer:
Given:
Wavelength of photon, λ = 228 Å
Energy
Eis given by
E=hcλHere, c = Speed of light
h = Planck’s constant
∴E=6.63×1034×3×108228×1010 =0.0872×1016 JAs the transition takes place from n = 1 to n = 2, the excitation energy (E_{1}) will be
E1=RhcZ21n121n22E1=13.6 eV×Z2×112122⇒E1= 13.6 eV×Z2×34This excitation energy should be equal to the energy of the photon.
∴13.6×34×Z2=0.0872×1016 Z2=0.0872×1016×413.6×3×1.6×1019=5.34 Z=5.34=2.3The ion may be helium.
Question 11:
Find the maximum Coulomb force that can act on the electron due to the nucleus in a hydrogen atom.
Answer:
Charge on the electron, q_{1} = 1.6
×10
19 C
Charge on the nucleus, q_{2} = 1.6
×10
19 C
Let r be the distance between the nucleus and the electron.
Coulomb force
Fis given by
F=q1q24π∈0r2 …..(1)
Here, q_{1} = q_{2} = q = 1.6
×10
19 C
Smallest distance between the nucleus and the first orbit, r = 0.53
×10
10 m
K=14πε0= 9
×10^{9} Nm^{2}C
2
Substituting the respective values in (1), we get
F=9×109×1.6×1019×1.6×10190.53×10102 =1.6×1.6×9×1090.532=82.02×109 =8.2×108 N
Question 12:
A hydrogen atom in a state having a binding energy of 0.85 eV makes transition to a state with excitation energy 10.2 e.V (a) Identify the quantum numbers n of the upper and the lower energy states involved in the transition. (b) Find the wavelength of the emitted radiation.
Answer:
(a)
The binding energy of hydrogen is given by
E=13.6n2eVFor binding energy of 0.85 eV,
n22=13.60.85=16n2=4For binding energy of 10.2 eV,
n12=13.610.2n1=1.15⇒n1=2The quantum number of the upper and the lower energy state are 4 and 2, respectively.
(b) Wavelength of the emitted radiation
λis given by
1λ=R1n121n22Here,
R = Rydberg constant
n_{1} and n_{2} are quantum numbers.
∴1λ=1.097×107 14116⇒λ=161.097×3×107 =4.8617×107 =487 nm
Question 13:
Whenever a photon is emitted by hydrogen in Balmer series, it is followed by another photon in Lyman series. What wavelength does this latter photon correspond to?
Answer:
As the second photon emitted lies in the Lyman series, the transition will be from the states having quantum numbers n = 2 to n = 1.
Wavelength of radiation
λis given by
1λ=R1n121n22Here, R is the Rydberg constant, having the value of 1.097×10^{7} m^{1}.
1λ=1.097×107 112122⇒1λ=1.097×107 114⇒ 1λ =1.097×34×107⇒λ=41.097×3×107 =1.215×107 =121.5×109=122 nm
Question 14:
A hydrogen atom in state n = 6 makes two successive transitions and reaches the ground state. In the first transition a photon of 1.13 eV is emitted. (a) Find the energy of the photon emitted in the second transition (b) What is the value of n in the intermediate state?
Answer:
Energy (E) of the nth state of hydrogen atom is given by
E =
13.6n2eV
For n = 6,
∴ E=13.636=0.377777777 eV
Energy of hydrogen atom in the ground state = −13.6 eV
Energy emitted in the second transition = Energy of ground state
– (Energy of hydrogen atom in the 6th state + Energy of photon)
=13.60.37777+1.13=12.09=12.1 eV(b)
Energy in the intermediate state = (Energy of photon emitted in the first transition) + (Energy of n = 6 state)
= 1.13 eV + 0.377 eV
= 1.507 eV
Energy of the nth state can expressed as
13.6n2=1.507
∴ n=13.61.507= 9.024≈3
Question 15:
What is the energy of a hydrogen atom in the first excited state if the potential energy is taken to be zero in the ground state?
Answer:
In ground state, the potential energy of a hydrogen atom is zero.
An electron is bound to the nucleus with an energy of 13.6 eV.
Therefore, we have to give 13.6 eV energy to move the electron from the nucleus.
Let us calculate the excitation energy required to take an atom from the ground state (n = 1) to the first excited state (n = 2).
E=13.6×1n121n22 eVTherefore, the excitation energy is given by
E=13.6×112122 eV E=13.6×34 eV=10.2 eVEnergy of 10.2 eV is needed to take an atom from the ground state to the first excited state.
∴ Total energy of an atom in the first excitation state = 13.6 eV + 10.2 eV = 23.8 eV
Question 16:
A hot gas emits radiation of wavelengths 46.0 nm, 82.8 nm and 103.5 nm only. Assume that the atoms have only two excited states and the difference between consecutive energy levels decreases as energy is increased. Taking the energy of the highest energy state to be zero, find the energies of the ground state and the first excited state.
Answer:
Given:
Energy (E) of the ground state will be the energy acquired in the transition of the 2 excitation state to ground state.
E1=hcλ1Here,
h = Planck’s constant
c = Speed of light
λ1 = Wavelength of the radiation emitted when atoms come from the highest excited state to ground state
∴ E1=(6.63×1034)×(3×108)(46×109) J E1=(6.63×1034)×(3×108)(46×109)×(1.6×1019) eV =124246=27 eVEnergy in the first excitation state
E2will be the energy acquired in the transition of the highest energy state to the 2nd excitation state.
E2=hcλnHere,
λn = Wavelength of the radiation emitted when an atom comes from the highest energy state to the 2nd excitation stateâ€‹.
E2=hcλnE2=(6.63×1034)×(3×108)(103.5×109) JE2=(6.63×1034)×(3×108)(103.5×109)×(1.6×1019) eV =12 eV
Question 17:
A gas of hydrogenlike ions is prepared in a particular excited state A. It emits photons having wavelength equal to the wavelength of the first line of the Lyman series together with photons of five other wavelengths. Identify the gas and find the principal quantum number of the state A.
Answer:
(a) If the atom is excited to the principal quantum (n), then the number of transitions is given by
nn12It is given that a total of 6 photons are emitted. Therefore, total number of transitions is 6.
∴
nn12 = 6
⇒ n = 4
Thus, the principal quantum number is 4 and the gas is in the 4th excited state.
Question 18:
Find the maximum angular speed of the electron of a hydrogen atom in a stationary orbit.
Answer:
Let the mass of the electron be m.
Let the radius of the hydrogen’s first stationary orbit be r.
Let the linear speed and the angular speed of the electron be v and ω, respectively.
According to the Bohr’s theory, angular momentum (L) of the electron is an integral multiple of h/2
π, where h is the Planck’s constant.
⇒mvr=nh2π(Here, n is an integer.)
v=rω⇒mr2ω=nh2π ⇒ω=nh2π×m×r2 ∴ ω = 1×6.63×10342×3.14×9.1093×1031×0.53×10102 = 0.413×1017 rad/s=4.13×1016 rad/s
Question 19:
A spectroscopic instrument can resolve two nearby wavelengths λ and λ + Δλ if λ/Δλ is smaller than 8000. This is used to study the spectral lines of the Balmer series of hydrogen. Approximately how many lines will be resolved by the instrument?
Answer:
The range of wavelength falling in Balmer series is between 656.3 nm and 365 nm.
It is given that the resolution of the instrument is λ/ (λ + âˆ†λ) < 8000.
Number of wavelengths in this range will be calculated in the following way:
656.33658000=36Two lines will be extra for the first and last wavelength.
∴ Total number of lines = 36 + 2 = 38
Question 20:
Suppose, in certain conditions only those transitions are allowed to hydrogen atoms in which the principal quantum number n changes by 2. (a) Find the smallest wavelength emitted by hydrogen. (b) List the wavelength emitted by hydrogen in the visible range (380 nm to 780 nm).
Answer:
Given:
Possible transitions:
From n_{1} = 1 to n_{2} = 3
n_{1} = 2 to n_{2} = 4
(a) Here, n_{1} = 1 and n_{2} = 3
Energy,
E=13.61n121n22
E=13.6 1119 =13.6×89 …..1Energy E is also given byE=hcλHere, h = Planck constant
c = Speed of the light
λ = Wavelength of the radiation
∴ E=6.63×1034×3×108λ …..2Equating equations 1 and 2, we have⇒λ=6.63×1034×3×108×913.6×8 =0.027×107=103 nm(b) Visible radiation comes in Balmer series.
As ‘n‘ changes by 2, we consider n = 2 to n = 4.
Energy,
E1=13.61n121n22
=13.6×14116 =2.55 eVIf
λ1 is the wavelength of the radiation, when transition takes place between quantum number n = 2 to n = 4, then
255 =
1242λ1or λ_{1}_{â€‹} = 487 nm
Question 21:
According to Maxwell’s theory of electrodynamics, an electron going in a circle should emit radiation of frequency equal to its frequency of revolution. What should be the wavelength of the radiation emitted by a hydrogen atom in ground state if this rule is followed?
Answer:
Let v_{0} be the velocity of the electron moving in the ground state and
r0be the radius of the ground state.
Frequency of the revolution of electron in the circle is given by
f=v02πr0Frequency of the radiation emitted = Frequency of the revolution of electron
∴ Frequency of the radiation emitted =
v02πr0 Also, c = f
λHere, c = Speed of light
λ = Wavelength of the radiation emitted
⇒ λ=cf
∴ λ=2πr0cv0 =2×3.14×53×1012×(3×108)2.187×106 =45.686×1012 m=45.7 nm
Page No 385:
Question 22:
The average kinetic energy of molecules in a gas at temperature T is 1.5 kT. Find the temperature at which the average kinetic energy of the molecules of hydrogen equals the binding energy of its atoms. Will hydrogen remain in molecular from at this temperature? Take k = 8.62 × 10^{−5} eV K^{−1}.
Answer:
Average kinetic energy
Kof the molecules in a gas at temperature
Tis given by
K =
32kTHere,
k = 8.62 × 10^{−5} eVK^{−1}
T = Temperature of gas
The binding energy of hydrogen atom is 13.6 eV.
According to the question,
Average kinetic energy of hydrogen molecules = Binding energy of hydrogen atom
∴1.5 kT = 13.6
⇒1.5 × 8.62 × 10^{−5} × T = 13.6
⇒T=13.61.5×8.62×105 =1.05×105 KNo, it is impossible for hydrogen to remain in molecular state at such a high temperature.
Question 23:
Find the temperature at which the average thermal kinetic energy is equal to the energy needed to take a hydrogen atom from its ground state to n = 3 state. Hydrogen can now emit red light of wavelength 653.1 nm. Because of Maxwellian distribution of speeds, a hydrogen sample emits red light at temperatures much lower than that obtained from this problem. Assume that hydrogen molecules dissociate into atoms.
Answer:
Given:
Wavelength of red light,
λ= 653.1 nm = 653.1
×10
9m
Kinetic energy of H_{2} molecules
Kis given by
K=32 kT …(1)
Here, k = 8.62 × 10^{−5} eV/K
T = Temperature of H_{2} molecules
Energy released
Ewhen atom goes from ground state to n = 3 is given by
E=13.61n121n22For ground state, n_{1} = 1
Also, n_{2} = 3
∴ E=13.6 1119 =13.6 89 …2Kinetic energy of H_{2} molecules = Energy released when hydrogen atom goes from ground state to n = 3 state
∴ 32×8.62×105 ×T=13.6×89⇒ T=13.6×8×29×3×8.62×105 =9.4×104 K
Question 24:
Average lifetime of a hydrogen atom excited to n = 2 state is 10^{−8} s. Find the number of revolutions made by the electron on the average before it jumps to the ground state.
Answer:
Frequency of electron (f) is given by
f=me44 ∈02 n3 h3Time period is given by
T=1fT=4 ∈02 n3 h3me4Here,
h = Planck’s constant
m = Mass of the electron
e = Charge on the electron
ε0= Permittivity of free space
∴ T=4×8.85×1012×23×6.63×103439.10×1031×1.6×10164 =12247.735 ×1019 sAverage life time of hydrogen, t = 10
8 s
Number of revolutions is given by
N=tT⇒N=10812247.735×1019N = 8.2
× 10^{5}^{ }revolution
Question 25:
Calculate the magnetic dipole moment corresponding to the motion of the electron in the ground state of a hydrogen atom.
Answer:
Mass of the electron, m = 9.1×10
31kg
Radius of the ground state, r = 0.53×10
10m
Let f be the frequency of revolution of the electron moving in ground state and A be the area of orbit.
Dipole moment of the electron (μ) is given by
μ = niA = qfA
=e×me44 ∈02 h3 n3×πr2 n2=me5×πr2n24 ∈02 h3 n3Here,
h = Planck’s constant
e = Charge on the electron
ε0 = Permittivity of free space
n = Principal quantum number
∴ μ=9.1×1031 1.6×10195×π×0.53×101024×8.85×10122×6.64×10343 ×13 =0.000917×1020 =9.176×1024 Am2
Question 26:
Show that the ratio of the magnetic dipole moment to the angular momentum (l = mvr) is a universal constant for hydrogenlike atoms and ions. Find its value.
Answer:
Mass of the electron, m = 9.1×10
31kg
Radius of the ground state, r = 0.53×10
10m
Let f be the frequency of revolution of the electron moving in the ground state and A be the area of orbit.
Dipole moment of the hydrogen like elements (μ) is given by
μ = niA = qfA
=e×me44 ∈02 h3 n3×πr02 n2=me5×πr02n24 ∈02 h3 n3Here,
h = Planck’s constant
e = Charge on the electron
ε0 = Permittivity of free space
n = Principal quantum number
Angular momentum of the electron in the hydrogen like atoms and ions (L) is given by
L=mvr=nh2πRatio of the dipole moment and the angular momentum is given by
μL=e5×m×πr2 n24∈0 h3 n3×2πnh
μL=1.6×10195×9.10×1031× 3.142×0.53×101022×8.85×10122×6.63×10343×12μL=3.73×1010 C/kgRatio of the magnetic dipole moment and the angular momentum do not depends on the atomic number ‘Z‘.
Hence, it is a universal constant.
Question 27:
A beam of light having wavelengths distributed uniformly between 450 nm to 550 nm passes through a sample of hydrogen gas. Which wavelength will have the least intensity in the transmitted beam?
Answer:
Given:
Minimum wavelength of the light component present in the beam,
λ1= 450 nm
Energy associated
E1with wavelength
λ1is given by
E_{1} =
hcλ1Here,
c = Speed of light
h = Planck’s constant
∴E1=1242450 = 2.76 eV
Maximum wavelength of the light component present in the beam,
λ2= 550 nm
Energy associated
E2with wavelength
λ2is given by
E_{2} =
hcλ2
∴ E2=1242550=2.228=2.26 eVThe given range of wavelengths lies in the visible range.
∴ n_{1} = 2, n_{2} = 3, 4, 5 …
Let E’_{2} , E’_{3} , E’_{4} and E’_{5} be the energies of the 2nd, 3rd, 4th and 5th states, respectively.
E’2E’3=13.61419 =12.6×530=1.9 eVE’2E’4=13.614116 =2.55 eVE’2E’5=13.614125 =10.5×21100=2.856 eVOnly, E’_{2} − E’_{4} comes in the range of the energy provided. So the wavelength of light having 2.55 eV will be absorbed.
λ=12422.55=487.05 nm =487 nmThe wavelength 487 nm will be absorbed by hydrogen gas. So, wavelength â€‹487 nm will have less intensity in the transmitted beam.
Question 28:
Radiation coming from transition n = 2 to n = 1 of hydrogen atoms falls on helium ions in n = 1 and n = 2 states. What are the possible transitions of helium ions as they absorbs energy from the radiation?
Answer:
Energy of radiation (E) from the hydrogen atom is given by
E=13.61n121n22Hydrogen atoms go through transition, n = 1 to n = 2.
The energy released is given by
E=13.61114 =13.6×34=10.2 eVFor He,
Atomic no, Z = 2
Let us check the energy required for the transition in helium ions from n = 1 to n = 2.
∴ n_{1} = 1 to n_{2} = 2
Energy
E1of this transition is given by
E1=Z213.61n121n22 =4×13.6 114 =40.8 eVE_{1} > E,
Hence, this transition of helium ions is not possible.
Let us check the energy required for the transition in helium ion from n = 1 to n = 3.
∴ n_{1} = 1 to n_{2} = 3
Energy
E2for this transition is given by
E2=Z2×13.6 1n221n12 =4×13.6×1119 =48.3 eVIt is clear that E_{2} > E.
Hence, this transition of helium ions is not possible.
Similarly, transition from n_{1} = 1 to n_{2} = 4 is also not possible.
Let us check the energy required for the transition in helium ion from n = 2 to n = 3.
∴ n_{1} = 2 to n_{2} = 3
Energy
E3for this transition is given by
E3=13.6×4 1419 =20×13.636=7.56 eVLet us check the energy required for the transition in helium ion from n = 2 to n = 3.
∴ n_{â€‹â€‹}_{1 }= 2 to n_{2} = 4
Energy
E4 for this transition is given by
E4=13.6×4 14116 =13.6×34=10.2 eVWe find that
E_{3} < E
E_{4} = E
Hence, possible transitions are from n = 2 to n = 3 and n = 2 to n = 4.
Question 29:
A hydrogen atom in ground state absorbs a photon of ultraviolet radiation of wavelength 50 nm. Assuming that the entire photon energy is taken up by the electron with what kinetic energy will the electron be ejected?
Answer:
Given:
Wavelength of ultraviolet radiation, λ = 50 nm = 50
×10
9m
We know that the work function of an atom is the energy required to remove an electron from the surface of the atom. So, we can find the work function by calculating the energy required to remove the electron from n_{1} = 1 to n_{2} = ∞.
Work function,
W0=13.6111∞ =13.6 eVUsing Einstein’s photoelectric equation, we get
E=W0+KE⇒hcλ13.6=KE ∵ E=hcλ⇒12425013.6=KE⇒KE=24.8413.6 =11.24 eV
Question 30:
A parallel beam of light of wavelength 100 nm passes through a sample of atomic hydrogen gas in ground state. (a) Assume that when a photon supplies some of its energy to a hydrogen atom, the rest of the energy appears as another photon. Neglecting the light emitted by the excited hydrogen atoms in the direction of the incident beam, what wavelengths may be observed in the transmitted beam? (b) A radiation detector is placed near the gas to detect radiation coming perpendicular to the incident beam. Find the wavelengths of radiation that may be detected by the detector.
Answer:
Given:
Wavelength of light, λ = 100 nm = 100
×10
9m
Energy of the incident light
Eis given by
E=hcλ
Here,
h = Planck’s constant
λ = Wavelength of light
∴E=1242100 E=12.42 eV(a)
Let E_{1}â€‹ and E_{2 }be the energies of the 1st and the 2nd state, respectively.
Let the transition take place from E_{1} to E_{2}.
Energy absorbed during this transition is calculated as follows:
Here,
n_{â€‹â€‹â€‹1} = 1
n_{â€‹â€‹â€‹2} = 2
Energy absorbed (E‘) is given by
E’=13.61n121n22 =13.61114 =13.6×34=10.2 eV Energy left = 12.42 eV − 10.2 eV = 2.22 eV
â€‹Energy of the photon =
hcλEquating the energy left with that of the photon, we get
2.22 eV=hcλ2.22 eV=1242λ or λ = 559.45 = 560 nm
Let E_{3} be the energy of the 3rd state.
Energy absorbed for the transition from E_{1} to E_{3} is given by
E’=13.61n121n22 =13.61119 =13.6×89=12.1 eVEnergy absorbed in the transition from E_{1} to E_{3} = 12.1 eV (Same as solved above)
Energy left = 12.42 − 12.1 = 0.32 eV
0.32=hcλ=1242λλ=12420.32 =3881.2=3881 nmLet E_{4} be the energy of the 4th state.
Energy absorbed in the transition from E_{3} to E_{4} is given by
E’=13.61n121n22 =13.619116 =13.6×7144=0.65 eVEnergy absorbed for the transition from n = 3 to n = 4 is 0.65 eV
Energy left = 12.42 − 0.65 = 11.77 eV
Equating this energy with the energy of the photon, we get
11.77=hcλor λ=124211.77=105.52The wavelengths observed in the transmitted beam are 105 nm, 560 nm and 3881 nm.
(b)
If the energy absorbed by the ‘H’ atom is radiated perpendicularly, then the wavelengths of the radiations detected are calculated in the following way:
E =10.2 eV⇒10.2=hcλor λ=124210.2=121.76 nm ≈121 nmE =12.1 eV⇒12.1=hcλor λ=124212.1=102.64 nm≈103 nmE =0.65 eV⇒0.65=hcλor λ=12420.65=1910.76 nm 1911 nmThus, the wavelengths of the radiations detected are 103 nm, 121 nm and 1911 nm.
Question 31:
A beam of monochromatic light of wavelength λ ejects photoelectrons from a cesium surface (Φ = 1.9 eV). These photoelectrons are made to collide with hydrogen atoms in ground state. Find the maximum value of λ for which (a) hydrogen atoms may be ionized, (b) hydrogen atoms may get excited from the ground state to the first excited state and (c) the excited hydrogen atoms may emit visible light.
Answer:
Given:
Work function of cesium surface, Ï• = 1.9 eV
(a) Energy required to ionise a hydrogen atom in its ground state, E = 13.6 eV
From the Einstein’s photoelectric equation,
hcλ=E+ϕHere,
h = Planck’s constant
c = Speed of light
λ= Wavelength of light
∴hcλ1.9=13.6⇒1240λ=15.5⇒λ=124015.5⇒λ = 80 nm(b) When the electron is excited from the states n_{1} = 1 to n_{2} = 2, energy absorbed
E1is given by
E1=13.61n121n22E1=13.6114E1=13.66×34For Einstein’s photoelectric equation,
∴hcλ1.9=13.6×34⇒hcλ=13.6×34+1.9⇒1240λ=10.2+1.9=12.1⇒λ=124012.1⇒ λ=102.47=102 nm(c) Excited atom will emit visible light if an electron jumps from the second orbit_{â€‹ }to third orbit, i.e. from n_{1} = 2â€‹ to n_{2} = 3. This is because Balmer series lies in the visible region.
Energy (E_{2}) of this transition is given by
E2 = 13.61n121n22E2 = 13.61419E2 = 13.66×536For Einstein’s photoelectric equation,
hcλ1.9 = 13.6×536⇒hcλ=13.6×536+1.9⇒1240λ = 1.88+1.9 = 3.78⇒λ=12403.78 ⇒ λ = 328.04 nm
Question 32:
Electrons are emitted from an electron gun at almost zero velocity and are accelerated by an electric field E through a distance of 1.0 m. The electrons are now scattered by an atomic hydrogen sample in ground state. What should be the minimum value of E so that red light of wavelength 656.3 nm may be emitted by the hydrogen?
Answer:
Given:
Distance travelled by the electron, d = 1.0 m
Wavelength of red light,
λ= 656.3 nm = 656.3
×10
9m
Since the given wavelength lies in Balmer series, the transition that requires minimum energy is from n_{1} = 3 to n_{2} = 2.
Energy of this transition will be equal to the energy
E1that will be required for the transition from the ground state to n = 3.
E1=13.61n121n22
⇒ E1=13.6119 =13.6×89=12.09 eVEnergy, E (eV) = 12.09 eV
∴ V = 12.09 V
Electric field, E =
Vd=
12.091= 12.09 V/m
∴ Minimum value of the electric field = 12.09 V/m = 12.1 V/m
Question 33:
A neutron having kinetic energy 12.5 eV collides with a hydrogen atom at rest. Nelgect the difference in mass between the neutron and the hydrogen atom and assume that the neutron does not leave its line of motion. Find the possible kinetic energies of the neutron after the event.
Answer:
Given:
Initial kinetic energy of the neutron, K = 12.5 eV
The velocities of the two bodies of equal masses undergoing elastic collision in one dimension gets interchanged after the collision.
Since the hydrogen atom is at rest, after collision, the velocity of the neutron will be zero.
Hence, it has zero energy.
Question 34:
A hydrogen atom moving at speed υ collides with another hydrogen atom kept at rest. Find the minimum value of υ for which one of the atoms may get ionized.
The mass of a hydrogen atom = 1.67 × 10^{−27} kg.
Answer:
Given:
Mass of the hydrogen atom, M = 1.67 × 10^{−27} kg
Let v be the velocity with which hydrogen atom is moving before collision.
Let v_{1} and v_{2} be the velocities of hydrogen atoms after the collision.
Energy used for the ionisation of one atom of hydrogen, ΔE = 13.6 eV = 13.6×(1.6×10^{−19})J
Applying the conservation of momentum, we get
mv = mv_{1} + mυ_{2} …(1)
Applying the conservation of mechanical energy, we get
12mυ2=12mυ12+12mυ22+∆E …2Using equation (1), we get
v^{2} = (v_{1} + v_{2})^{2}
v^{2}
=υ12+υ22+2υ1υ2 …(3)
In equation(2), on multiplying both the sides by 2 and dividing both the sides by m.
υ2=υ12+υ12+2∆E/m ….4On comparing (4) and (3), we get
∵2υ1υ2=2∆Em(υ_{1} − υ_{2})^{2} = (υ_{1} + υ_{2})^{2} − 4υ_{1}υ_{2}
_{â€‹}
v1v22=v24∆Em
For minimum value of υ,
v_{1}_{ }= v_{2}_{ }=0
Also,
v24∆Em=0
∴ v2=4∆Em =4×13.6×1.6×10191.67×1027 v=4×13.6×1.6×10191.67×1027 =1044×13.6×1.61.67 =7.2×104 m/s
Question 35:
A neutron moving with a speed υ strikes a hydrogen atom in ground state moving towards it with the same speed. Find the minimum speed of the neutron for which inelastic (completely or partially) collision may take place. The mass of neutron = mass of hydrogen = 1.67 × 10^{−27} kg.
Answer:
Given:
Mass of neutron, m = 1.67 × 10^{−27} kg
Since neutron is moving with velocity
v, its energy
Eis given by
E=12mυ2Let the energy absorbed be âˆ†E.
The condition for inelastic collision is given below:
12mυ2>2∆E
⇒∆E<14mυ2Since 10.2 eV, energy is required for the first excited state.
∴ âˆ†E < 10.2 eV
∴10.2 eV<14mv2Thus, minimum speed of the neutron is given by
⇒υmin=4×10.2meV⇒υmin=10.2×1.6×1019×41.67×1027 =6×104 m/sec
Question 36:
When a photon is emitted by a hydrogen atom, the photon carries a momentum with it. (a) Calculate the momentum carries by the photon when a hydrogen atom emits light of wavelength 656.3 nm. (b) With what speed does the atom recoil during this transition? Take the mass of the hydrogen atom = 1.67 × 10^{−27} kg. (c) Find the kinetic energy of recoil of the atom.
Answer:
Given:
Wavelength of light emitted by hydrogen, λ = 656.3 nm
Mass of hydrogen atom, m = 1.67 × 10^{−27} kg
(a) Momentum
pis given by
P=hλHere,
h = Planck’s constant
λ = Wavelength of light
∴ p =6.63×1034656.3×109 p =0.01×1025 p =1×1027 kgm/s(b) Momentum, p = mv
Here,
m = Mass of hydrogen atom
v = Speed of atom
∴1 × 10^{−27} = (1.67 × 10^{−27})× υ
⇒υ=11.67 =0.598=0.6 m/s(c) Kinetic energy
Kof the recoil of the atom is given by
K=12mv2 Here,
m = Mass of the atom
v = Velocity of the atom
∴ K=12×1.67×1027×0.62 J
K=0.3006×10271.6×1019 eVK=1.9×109 eV
Question 37:
When a photon is emitted from an atom, the atom recoils. The kinetic energy of recoil and the energy of the photon come from the difference in energies between the states involved in the transition. Suppose, a hydrogen atom changes its state from n = 3 to n = 2. Calculate the fractional change in the wavelength of light emitted, due to the recoil.
Answer:
Difference in energy in the transition from n = 3 to n = 2 is 1.89 eV ( = E).
If all this energy is used up in emitting a photon (i.e. recoil energy is zero).
Then,
E=hcλ ⇒λ=hcE …(i)If difference of energy is used up in emitting a photon and recoil of atom, then let E_{R} be the recoil energy of atom.
E=hcλ’+ER⇒λ’=hcEER …(ii)Fractional change in the wavelength is given as,
∆λλ=λ’λλ⇒∆λλ=1λhcEERhcE⇒∆λλ=EhchcERE(EER) ∵λ=hcE⇒∆λλ=ER(EER
Question 38:
The light emitted in the transition n = 3 to n = 2 in hydrogen is called H_{α} light. Find the maximum work function a metal can have so that H_{α} light can emit photoelectrons from it.
Answer:
The H_{α} light can emit the photoelectrons if its energy is greater than or equal to the work function of the metal.
Energy possessed by H_{α} light (E) is given by
E=13.61n121n22eV
Here, n_{1} = 2, n_{2} = 3â€‹
∴ E=13.6×1419 =13.6×536=1.89 eV = 1.90 eV H_{α} light will be able to emit electron from the metal surface for the maximum work function of metal to be 1.90 eV.
Question 39:
Light from Balmer series of hydrogen is able to eject photoelectrons from a metal. What can be the maximum work function of the metal?
Answer:
Let the maximum work function of the metal be W.
The energy liberated in the Balmer Series (E) is given by
E=13.61n121n22For maximum work function, maximum energy of Balmer’s series is taken.
Now, n_{1} = 2, n_{1} = ∞â€‹
∴ E=13.6122 =13.6×14=3.4 eVHere,
W = E
Thus, maximum work function of metal is 3.4 eV.
Question 40:
Radiation from hydrogen discharge tube falls on a cesium plate. Find the maximum possible kinetic energy of the photoelectrons. Work function of cesium is 1.9 eV.
Answer:
Given:
Work function of cesium,
ϕ = 1.9 eVâ€‹
Energy of photons coming from the discharge tube, E = 13.6 eV
Let maximum kinetic energy of photoelectrons emitted be K.
From the Einstein’s photoelectric equation, we know that the maximum kinetic energy of photoelectrons emitted is given by
K = E −
ϕ = 13.6 eV − 1.9 ev
= 11.7 eVâ€‹
Question 41:
A filter transmits only the radiation of wavelength greater than 440 nm. Radiation from a hydrogendischarge tube goes through such a filter and is incident on a metal of work function 2.0 eV. Find the stopping potential which can stop the photoelectrons.
Answer:
Wavelength of radiation coming from filter, λ = 440 nm
Work function of metal, Ï• = 2 eV
Charge of the electronâ€‹, e = 1.6
× 10
9 C
Let V_{0} be the stopping potential.
From Einstein’s photoelectric equation,
hcλϕ=eV0Here,
h =Planck constant
c = Speed of light
λ = Wavelength of radiation
4.14×1015×3×108440×1092 eV=eV0⇒eV0=12424402 eV=0.823 eV.⇒V0=0.823 Volts
Question 42:
The earth revolves round the sun due to gravitational attraction. Suppose that the sun and the earth are point particles with their existing masses and that Bohr’s quantization rule for angular momentum is valid in the case of gravitation. (a) Calculate the minimum radius the earth can have for its orbit. (b) What is the value of the principal quantum number n for the present radius? Mass of the earth = 6.0 × 10^{−24} kg. Mass of the sun = 2.0 × 10^{30} kg, earthsun distance = 1.5 × 10^{11} m.
Answer:
Given:
Mass of the earth, m_{e} = 6.0 × 10^{24} kg
Mass of the sun, m_{s} = 2.0 × 10^{30} kg
Distance between the earth and the sun, d = 1.5 × 11^{11} m
According to the Bohr’s quantization rule,
Angular momentum, L =
nh2π
⇒ mvr=nh2π ….(1)
Here,
n = Quantum number
h = Planck’s constant
m = Mass of electron
r = Radius of the circular orbit
v = Velocity of the electron
Squaring both the sides, we get
me2v2r2=n2h24π2 ….2Gravitational force of attraction between the earth and the sun acts as the centripetal force.
F=Gmemsr2=mev2r
⇒v2=Gmsr ….(3)
Dividing (2) by (3), we get
me2 r=n2h24π2Gms(a) For n = 1,
r=h24π2 Gms me2r =6.63×103424×3.142×6.67×1011×6×10242×2×1030r =2.29×10138 mr =2.3×10138 m
(b)
From (2), the value of the principal quantum number (n) is given by
n2=me2×r×4×π×G×msh2⇒n=me2×r×4×π×G×msh2
n=6×10242×1.5×1011×4×3.142×6.67×1011×2×10306.6×10342n=2.5×1074
Page No 386:
Question 43:
Consider a neutron and an electron bound to each other due to gravitational force. Assuming Bohr’s quantization rule for angular momentum to be valid in this case, derive an expression for the energy of the neutronelectron system.
Answer:
According to Bohr’s quantization rule,
Angular momentum of electron, L =
nh2π
⇒mver=nh2π⇒ve=nh2πrme …(1)
Here,
n = Quantum number
h = Planck’s constant
m = Mass of the electron
r = Radius of the circular orbit
v_{e} = Velocity of the electron
Let m_{n} be the mass of neutron.
On equating the gravitational force between neutron and electron with the centripetal acceleration,
Gmnmer2=mev2r⇒Gmnr=v2 …2Squaring (1) and dividing it by (2), we have
me2v2r2v2=n2h2r4π2Gmn⇒me2r2=n2h2r4π2Gmn⇒r=n2h24π2Gmnme2ve=nh2πrme⇒ve=nh2πme×n2h24π2Gmnme2⇒ve=2πGmnmenh Kinetic energy of the electron, K=12meve2 =12me2πGmnmenh2 =4π2G2mn2me32n2h2Potential energy of the neutron,
P=GmnmerSubstituting the value of r in the above expression,
P=Gmemn4π2Gmnme2n2h2P=4π2G2mn2me3n2h2Total energy = K + P
=2π2G2mn2me22n2h2=π2G2mn2me2n2h2
Question 44:
A uniform magnetic field B exist in a region. An electron projected perpendicular to the field goes in a circle. Assuming Bohr’s quantization rule for angular momentum, calculate (a) the smallest possible radius of the electron (b) the radius of the nth orbit and (c) the minimum possible speed of the electron.
Answer:
According to Bohr’s quantization rule,
mvr = nh2π’r‘ is minimum when ‘n‘ has minimum value, i.e. 1.
mv=nh2πr …1Again, r=mvqB⇒ mv=rqB …2From (1) and (2), we get
rqB=nh2πr From 1⇒r2=nh2πeB ∴q=e⇒r=h2πeB n=1(b) For the radius of n^{th} orbit,
r=nh2πeB(c)
mvr=nh2π, r=mvqBSubstituting the value of ‘r‘ in (1), we get
mv×mvqB=nh2π⇒m2v2=heB2π n=1, q=e⇒v2=heB2πm2 v=heB2πm2
Question 45:
Suppose in an imaginary world the angular momentum is quantized to be even integral multiples of h/2π. What is the longest possible wavelength emitted by hydrogen atoms in visible range in such a world according to Bohr’s model?
Answer:
In the imaginary world, the angular momentum is quantized to be an even integral multiple of h/2
π.
Therefore, the quantum numbers that are allowed are n_{1} = 2 and n_{2} = 4.
We have the longest possible wavelength for minimum energy.
Energy of the light emitted (E) is given by
E=13.61n121n22E=13.6122142E=13.614116E=13.6×1264=2.55 eVEquating the calculated energy with that of photon, we get
2.55 eV=hcλλ=hc2.55=12422.55 nm =487.05 nm=487 nm
Question 46:
Consider an excited hydrogen atom in state n moving with a velocity υ(ν<<c). It emits a photon in the direction of its motion and changes its state to a lower state m. Apply momentum and energy conservation principles to calculate the frequency ν of the emitted radiation. Compare this with the frequency ν_{0} emitted if the atom were at rest.
Answer:
Let the frequency emitted by the atom at rest be ν_{0}.
Let the velocity of hydrogen atom in state ‘n‘ be u.
But u << c
Here, the velocity of the emitted photon must be u.
According to the Doppler’s effect,
The frequency of the emitted radiation, ν is given by
Frequency of the emitted radiation, ν=ν01+uc1ucSince u <<< c,ν=ν0 1+uc1 ν=ν01+ucRatio of frequencies of the emitted radiation,
νν0=1+uc
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