Hc Verma II for Class 11 Science Physics Chapter 43 - Bohr S Model And Physics Of The Atom

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Page No 382:

Answer:

Balmer series contains wavelengths ranging from 364 nm (for n₂ = 3) to 655 nm (n₂= $\infty$ ).
So, the given range of wavelength (380−780 nm) lies in the Balmer series.
The wavelength in the Balmer series can be found by
$\frac{1}{λ} = R (\frac{1}{2^{2}} - \frac{1}{n^{2}})$
Here, R = Rydberg's constant = 1.097×10⁷ m^{$-$ 1}

The wavelength for the transition from n = 3 to n = 2 is given by
$\frac{1}{λ_{1}} = R (\frac{1}{2^{2}} - \frac{1}{3^{2}}) λ_{1} = 656.3 nm$
The wavelength for the transition from n = 4 to n = 2 is given by
$\frac{1}{λ_{2}} = R (\frac{1}{2^{2}} - \frac{1}{4^{2}}) λ_{2} = 486.1 nm$
The wavelength for the transition from n = 5 to n = 2 is given by
$\frac{1}{λ_{3}} = R (\frac{1}{2^{2}} - \frac{1}{5^{2}}) λ_{3} = 434.0 nm$
The wavelength for the transition from n = 6 to n = 2 is given by
$\frac{1}{λ_{4}} = R (\frac{1}{2^{2}} - \frac{1}{6^{2}}) λ_{4} = 410.2 nm$
The wavelength for the transition from n = 7 to n = 2 is given by
$\frac{1}{λ_{5}} = R (\frac{1}{2^{2}} - \frac{1}{7^{2}}) λ_{5} = 397.0 nm$

Thus, the wavelengths emitted by the atomic hydrogen in visible range (380−780 nm) are 5.
Lyman series contains wavelengths ranging from 91 nm (for n₂ = 2) to 121 nm (n₂= $\infty$ ).
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
$\frac{1}{λ} = R (\frac{1}{1^{2}} - \frac{1}{n^{2}})$

The wavelength for the transition from n = 2 to n = 1 is given by
$\frac{1}{λ_{1}} = R (\frac{1}{1^{2}} - \frac{1}{2^{2}}) λ_{1} = 122 nm$
The wavelength for the transition from n = 3 to n = 1 is given by
$\frac{1}{λ_{2}} = R (\frac{1}{1^{2}} - \frac{1}{2^{2}}) λ_{2} = 103 nm$
The wavelength for the transition from n = 4 to n = 1 is given by
$\frac{1}{λ_{3}} = R (\frac{1}{1^{2}} - \frac{1}{4^{2}}) λ_{3} = 97.3 nm$
The wavelength for the transition from n = 5 to n = 1 is given by
$\frac{1}{λ_{4}} = R (\frac{1}{1^{2}} - \frac{1}{5^{2}}) λ_{4} = 95.0 nm$
The wavelength for the transition from n = 6 to n = 1 is given by
$\frac{1}{λ_{5}} = R (\frac{1}{1^{2}} - \frac{1}{6^{2}}) λ_{5} = 93.8 nm$

So, it can be noted that the number of wavelengths lying between 50 nm to 100 nm are 3.

Page No 382:

Answer:

The energy of hydrogen ion is given by
$E_{n} = - \frac{(13.6 eV) Z^{2}}{n^{2}}$
For 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.

Page No 382:

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
$E = (13.6 eV) \times (\frac{1}{1^{2}} - \frac{1}{n^{2}})$
Substituting 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).

Page No 382:

Answer:

White radiations are x-rays that have energy ranging between 5-10 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:

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 n₂ = $\infty$ ) to 655 nm (for n₂=3).

Page No 383:

Answer:

Energy of n^th state of hydrogen is given by
$E_{n} = \frac{- 13.6}{n^{2}} eV$
Energy of first excited state (n = 2) of hydrogen, E₁ = $\frac{- 13.6}{4} eV$ = -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,
E₁^, = -3.4 eV-10 eV = -13.4 eV
We still write E_n = E₁/n², or r_n = a₀ n² because these formulas are independent of the refrence point enegy.

Page No 383:

Answer:

The 'series limit' refers to the 'shortest wavelength' (corresponding to the maximum photon energy).

The frequency of the radiation emitted for transition from n₁ to n₂is given by
$f = k (\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}})$
Here, k is a constant.

For the series limit of Lyman series,
n₁ = 1
n₂ = $\infty$

Frequency, $f_{1} = k (\frac{1}{1^{2}} - \frac{1}{\infty}) = k$

For the first line of Lyman series,
n₁ = 1
n₂ = 2

Frequency, $f_{2} = k (\frac{1}{1^{2}} - \frac{1}{2^{2}}) = \frac{3 k}{4}$

For series limit of Balmer series,
n₁ = 2
n₂ = $\infty$

$f_{1} = k (\frac{1}{2^{2}} - \frac{1}{\infty}) = \frac{k}{4}$

f₁ $-$ f₃= f₂

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.

Page No 383:

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.

Page No 383:

Answer:

When a photon of energy (E₂ – E₁ = hυ) is incident on an atom in the ground state, the atom in the ground state E₁ may absorb the photon and jump to a higher energy state (E₂). 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.

Page No 383:

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
$R = \frac{A}{B ρ (ν)} = e^{\frac{h ν}{k T}} - 1$
For microwave region
$ν = 10^{10} (say) R = e^{\frac{6.6 \times 10^{- 34} \times 10^{10}}{1.38 \times 10^{- 23} \times 300}} - 1 = e^{0.0016} - 1 = 0.0016$
This implies that stimulated transition dominate in this region.
For visible region
$ν = 10^{15} \Rightarrow R = e^{160} - 1 \Rightarrow R > > 1$
So here spontaneous transition dominate.

Page No 383:

Answer:

(c) h/2π

According to Bohr's atomic theory, the orbital angular momentum of an electron is an integral multiplt of h/2π.
∴ $L_{n} = \frac{n h}{2 π}$

Here,
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
$L = \frac{h}{2 π}$

Page No 383:

Answer:

(d) either two atoms or three atoms

The energies of the photons emitted can be expressed as follows:
$13.6 (\frac{1}{1^{2}} - \frac{1}{2^{2}}) eV = 10.2 eV$
$13.6 (\frac{1}{1^{2}} - \frac{1}{3^{2}}) eV = 12.1 eV$
$13.6 (\frac{1}{2^{2}} - \frac{1}{3^{2}}) eV = 1.9 eV$

The 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).

Page No 383:

Answer:

(c) 10⁻⁴²N-m

The angular momentum of the electron for the nth state is given by
$L_{n} = \frac{n h}{2 π}$
Angular momentum of the electron for n = 3, $L_{i} = \frac{3 h}{2 π}$

Angular momentum of the electron for n = 2, $L_{f} = \frac{2 h}{2 π}$

The torque is the time rate of change of the angular momentum.
$Torque, τ = \frac{L_{f} - L_{i}}{t} = \frac{(2 h / 2 π) - (3 h / 2 π)}{10^{- 8}} = \frac{- (h / 2 π)}{10^{- 8}} = \frac{- 10^{- 34}}{10^{- 8}} [∵ \frac{h}{2 π} \approx 10^{- 34} J - s] = - 10^{- 42} N - m$
The magnitude of the torque is 10^{$-$ 42}N-m.

Page No 383:

Answer:

(d) n = 2 to n = 1

For the transition in the hydrogen-like atom, the wavelength of the emitted radiation is calculated by

$\frac{1}{λ} = R Z^{2} (\frac{1}{n_{1}} - \frac{1}{n_{2}})$
Here, R is the Rydberg constant.

For the transition from n = 5 to n = 4, the wavelength is given by
$\frac{1}{λ} = R Z^{2} (\frac{1}{4^{2}} - \frac{1}{5^{2}}) λ = \frac{400}{9 R Z^{2}}$

For the transition from n = 4 to n = 3, the wavelength is given by
$\frac{1}{λ} = R Z^{2} (\frac{1}{3^{2}} - \frac{1}{4^{2}}) λ = \frac{144}{7 R Z^{2}}$

For the transition from n = 3 to n = 2, the wavelength is given by
$\frac{1}{λ} = R Z^{2} (\frac{1}{2^{2}} - \frac{1}{3^{2}}) λ = \frac{36}{5 R Z^{2}}$

For the transition from n = 2 to n = 1, the wavelength is given by
$\frac{1}{λ} = R Z^{2} (\frac{1}{1^{2}} - \frac{1}{2^{2}}) λ = \frac{2}{R Z^{2}}$

From 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.

Page No 383:

Answer:

(d) Doubly ionized lithium

For a hydrogen-like ion with Z protons in the nucleus, the radius of the nth state is given by

$r_{n} = \frac{n^{2} a_{0}}{Z}$

Here, a₀ = 0.53 pm

For lithium,
Z = 3
Therefore, the radius of the first orbit for doubly ionised lithium will be minimum.

Page No 383:

Answer:

(d) Doubly ionized lithium

The wavelength corresponding the transition from n₂ to n₁ is given by
$\frac{1}{λ} = R Z^{2} (\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}})$
Here,
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 383:

Answer:

(c)

The speed (v) of electron can be expressed as
$v = \frac{Z e^{2}}{2 \in_{0} h n}$ ....(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).

Page No 383:

Answer:

(b) increases

The electric potential energy of hydrogen atom with electron at the nth state is given by
$V = - \frac{2 \times 13.6}{n^{2}}$
As the value of n increases, the potential energy of the hydrogen atom also increases, i.e. the atom becomes less bound as n increases.

Page No 383:

Answer:

(c) He⁺

The total energy of a hydrogen-like ion, having Z protons in its nucleus, is given by
$E = - \frac{13.6 Z^{2}}{n^{2}}$ eV

Here, n = Principal quantum number

For ground state,
n = 1
∴ Total energy, E = $-$ 13.6 Z² 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 = $-$ 13.6×2² = $-$ 54.4 eV

For Li⁺⁺,
Z = 3
∴ Total energy, E = $-$ 13.6×3² = $-$ 122.4 eV

Hence, the ion having an energy of −54.4 eV in its ground state may be He⁺.

Page No 383:

Answer:

(d) Li⁺⁺

The radius of the nth orbit in one electron system is given by
$r_{n} = \frac{n^{2} a_{0}}{Z}$

Here, a₀ = 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 = $\frac{53}{2} pm = 26.5 pm$

For Li⁺⁺,
Z = 3

∴ Radius of Li⁺⁺ atom = $\frac{53}{3} pm = 17.66 pm \approx 18 pm$

The given one-electron system having radius of the shortest orbit to be 18 pm may be Li⁺⁺.

Page No 383:

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
$10.2 = 13.6 \times (\frac{1}{1^{2}} - \frac{1}{n^{2}}) \Rightarrow \frac{10.2}{13.6} = 1 - \frac{1}{n^{2}} \Rightarrow \frac{1}{n^{2}} = \frac{13.6 - 10.2}{13.6} \Rightarrow \frac{1}{n^{2}} = \frac{3.4}{13.6} \Rightarrow n^{2} = 4 \Rightarrow n = 2$

The orbital angular momentum of the electron in the nth state is given by
$L_{n} = \frac{n h}{2 π}$
Change in the angular momentum, $∆ L = \frac{2 h}{2 π} - \frac{h}{2 π} = \frac{h}{2 π}$
∴ $∆ L = 1.05 \times 10^{- 34} Js$

Page No 383:

Answer:

(d) Orbital angular momentum of the electron

According to Bohr's atomic theory, the orbital angular momentum of an electron in a one-electron system is given by
$L_{n} = \frac{n h}{2 π}$

Here,
n = Principal quantum number

The angular momentum is independent of the atomic number of the one-electron system. Therefore, it is same for all hydrogen-like atoms and ions in their ground states.
The other parameters given here are dependent on the atomic number of the hydrogen-like atom or ion taken.

Page No 383:

Answer:

(d) move with same speed

All the photons emitted in the laser move with the speed equal to the speed of light (c = 3×10⁸ 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.

Page No 383:

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:

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.

Page No 384:

Answer:

(a) vn
(b) Er
Relations for energy, radius of the orbit and its velocity are given by
$E = - \frac{m Z^{2} e^{4}}{8 {\in_{0}}^{2} h^{2} n^{2}} r = \frac{\in_{0} h^{2} n^{2}}{{πmZe}^{2}} v = \frac{Z e^{2}}{2 \in_{0} h n}$
Where
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
$\in_{0}$ : permittivity of vacuum
From these relations, we can see that the products independent of n are vn, Er.

Page No 384:

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
$r_{n} = n^{2} a_{0}$
Area of the nth orbit is given by
$A_{n} = π r_{n}^{2} = π n^{4} a_{0}^{2} A_{1} = π a_{0}^{2} \Rightarrow \ln (\frac{A_{n}}{A_{1}}) = \ln (\frac{π n^{4} a_{0}^{2}}{π a_{0}^{2}}) \ln (\frac{A_{n}}{A_{1}}) = 4 \ln n . . . (1)$
From the above expression, the graph of ln (A_n/A₁) against ln(n) will be a straight line passing through the origin and having slope 4.

Page No 384:

Answer:

(b) u_A > u_B
The ionisation energy of a hydrogen like ion of atomic number Z is given by
$V = (13.6 eV) \times Z^{2}$

Thus, 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_Bis correct.
The total energy of the atom is given by
$E = - \frac{m Z^{2} e^{2}}{8 \in_{0} h^{2} n^{2}}$
As the energy is directly proportional to Z², 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_Bis invalid.

Page No 384:

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.

Page No 384:

Answer:

The dimensions of ε₀ can be derived from the formula given below:

$a = \frac{ε_{0} h^{2}}{π m e^{2}} = \frac{A^{2} T^{2} {({ML}^{2} T^{- 1})}^{2}}{L^{2} {ML}^{- 2} M {(AT)}^{2}} = \frac{M^{2} L^{2} T^{- 2}}{M^{2} L^{3} T^{- 2}} = L$

Clearly, a₀ has the dimensions of length.

Page No 384:

Answer:

From Balmer empirical formula, the wavelength $(λ)$ of the radiation is given by
$\frac{1}{λ} = R (\frac{1}{{n_{1}}^{2}} - \frac{1}{{n_{2}}^{2}})$
Here, R = Rydberg constant = 1.097 $\times$ 10⁷m^{$- 1$}
n₁ = Quantum number of final state
n₂ = Quantum number of initial state
(a)
For transition from n = 3 to n = 2:

Here,
n₁ = 2
n₂ = 3

$\frac{1}{λ} = 1.09737 \times 10^{7} \times (\frac{1}{4} - \frac{1}{9}) \Rightarrow λ = \frac{36}{5 \times 1.09737 \times 10^{7}} = 6.56 \times 10^{- 7} = 656 nm$

(b)
For transition from n = 5 to n = 4:

Here,
n₁ = 4
n₂ = 5

$\frac{1}{λ} = 1.09737 \times 10^{- 7} (\frac{1}{16} - \frac{1}{25}) \Rightarrow λ = \frac{400}{1.09737 \times 10^{7} \times 9} = 4050 nm$

(c)
For transition from n = 10 to n = 9:

Here,
n₁ = 9
n₂ = 10

$\frac{1}{λ} = 1.09737 \times 10^{7} (\frac{1}{81} - \frac{1}{100}) λ = \frac{81 \times 100}{19 \times 1.09737 \times 10^{7}} = 38849 nm$

Page No 384:

Answer:

Given:
For the smallest wavelength, energy should be maximum.
Thus, for maximum energy, transition should be from infinity to the ground state.

∴ n₁= 1
n₂= $\infty$

(a) Wavelength of the radiation emitted $(λ)$ is given by
$\frac{1}{λ} = R Z^{2} (\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}})$

For hydrogen,
Atomic number, Z = 1
R = Rydberg constant = 1.097×10⁷ m^{$-$ 1}

On substituting the respective values,
$λ = \frac{1}{1.097 \times 10^{7}} = \frac{1}{1.097} \times 10^{- 7} = 0.911 \times 10^{- 7} = 91.1 \times 10^{- 9} = 91 nm$

(b)
For He⁺,
Atomic number, Z = 2
Wavelength of the radiation emitted by He⁺( $λ$ ) is given by
$\frac{1}{λ} = R Z^{2} (\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}})$
$∴ \frac{1}{λ} = {(2)}^{2} (1.097 \times 10^{7}) (\frac{1}{{(1)}^{2}} - \frac{1}{{(\infty)}^{2}})$
$\Rightarrow λ = \frac{91 nm}{4} = 23 nm$

(c) For Li⁺⁺,
Atomic number, Z = 3
Wavelength of the radiation emitted by Li⁺⁺( $λ$ ) is given by
$\frac{1}{λ} = R Z^{2} (\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}})$
$∴ \frac{1}{λ} = {(3)}^{2} \times (1.097 \times 10^{7}) (\frac{1}{1^{2}} - \frac{1}{\infty^{2}})$
$\Rightarrow λ = \frac{91 nm}{Z^{2}} = \frac{91}{9} = 10 nm$

Page No 384:

Answer:

Expression of Rydberg constant (R) is given by
$R = \frac{m e^{4}}{8 h^{2} c \in_{0}^{2}}$
Mass of electron, m_e = 9.31 $\times$ 10²¹ kg
Charge, e = 1.6 × 10⁻¹⁹ C
Planck's constant, h = 6.63 × 10⁻³⁴ J-s,
Speed of light, c = 3 × 10⁸ m/s,
Permittivity of vacuum, ∈₀ = 8.85 × 10⁻¹²C²N^{$- 1$}m
On substituting the values in the expression, we get
^{$R = \frac{(9.31 \times 10^{- 31}) \times {(1.6 \times 10^{- 19})}^{4}}{8 \times {(6.63 \times 10^{- 34})}^{2} \times (3 \times 10^{8}) \times {(8.85 \times 10^{- 12})}^{2}} \Rightarrow R = 1.097 \times 10^{7} m^{- 1}$}

Page No 384:

Answer:

The binding energy (E) of hydrogen atom is given by
$E = \frac{- 13.6}{n^{2}} eV$
For state n = 2,
$E = - \frac{13.6}{{(2)}^{2}} \Rightarrow E = - 3.4 eV$

Thus, binding energy of hydrogen at n = 2 is $- 3.4 eV$ .

Page No 384:

Answer:

For He⁺ ion,
Atomic number, Z = 2
For hydrogen like ions, radius $(r)$ of the nth state is given by
$r = \frac{0.53 n^{2}}{Z}$ Å
Here,
Z = Atomic number of ions
n = Quantum number of the state

Energy $(E)$ of the nth state is given by
E_n = $- \frac{13.6 Z^{2}}{n^{2}}$
(a)
For n = 1,
Radius, $r = \frac{0.53 \times {(1)}^{2}}{2} = 0.265 A^{0}$ Å

Energy, E_n = $\frac{- 13.6 \times 4}{1}$
= $-$ 54.4 eV

(b)
For n = 4,
Radius, $r = \frac{0.53 \times 16}{2} = 4.24 Å$
Energy, $E = \frac{- 13.6 \times 4}{16} = - 3.4 e V$
(c)
For n = 10,
Radius, $r = \frac{0.53 \times 100}{2}$
= 26.5 Å
Energy, $E = \frac{- 13.6 \times 4}{100} = - 0.544 eV$

Page No 384:

Answer:

Wavelength of ultraviolet radiation, $λ$ = 102.5 nm = 102,5 $\times$ 10^{$- 9$} m
Rydberg's constant, R = 1.097 $\times$ 10⁷ 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 (n₁= 1) from any excited state (n₂).
Wavelength of light $(λ)$ is given by
$\frac{1}{λ} = R (\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}) Here, R = Rydberg constant \Rightarrow \frac{1}{102.5 \times 10^{- 9}} = 1.097 \times 10^{7} (\frac{1}{1^{2}} - \frac{1}{n_{2}^{2}}) \Rightarrow \frac{10^{9}}{102.5} = 1.097 \times 10^{7} (1 - \frac{1}{n_{2}^{2}}) \Rightarrow \frac{10^{2}}{102.5} = 1.097 (1 - \frac{1}{n_{2}^{2}}) \Rightarrow 1 - \frac{1}{n_{2}^{2}} = \frac{100}{102.5 \times 1.097} \Rightarrow \frac{1}{n_{2}^{2}} = 1 - \frac{100}{102.5 \times 1.097} \Rightarrow n_{2} = \sqrt{9.041} = 3$
The transition will be from 1 to 3.

Page No 384:

Answer:

(a) PE of hydrogen like atom in the nth state, V = $= \frac{- 13.6 Z^{2}}{n^{2}} eV$
Here, 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⁺
$- 13.6 Z^{2} (1 - \frac{1}{2^{2}}) e V = - 10.2 Z^{2} e V$
⇒10.2 × Z²
= 10.2 × 4
= 40.8 V

(b) Ionization Potential Li⁺⁺ = 13.6 V × Z²
= 13.6 × 9
= 122.4 V

Page No 384:

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.
Here,
n₁= 2
n₂= 4
Now, the wavelength ( $λ$ ₁) will be
$\frac{1}{λ_{1}} = R (\frac{1}{{n_{1}}^{2}} - \frac{1}{{n_{2}}^{2}})$
$R = 1.097 \times 10^{7} m^{- 1} \frac{1}{λ_{1}} = 1.097 \times 10^{7} \times (\frac{1}{4} - \frac{1}{16}) \Rightarrow \frac{1}{λ_{1}} = 1.097 \times 10^{7} (\frac{4 - 1}{16}) \Rightarrow \frac{1}{λ_{1}} = \frac{1.097 \times 10^{7} \times 3}{16} \Rightarrow λ_{1} = \frac{16 \times 10^{- 7}}{3 \times 1.097} = 4.8617 \times 10^{- 7} = 486.1 \times 10^{- 9} = 487 nm$

When an atom makes transition from (n = 4) to (n = 3), the wavelength ( $λ$ 2) is given by

$Here again, n_{1} = 3 n_{2} = 4 \frac{1}{λ_{2}} = 1.097 \times 10^{7} (\frac{1}{9} - \frac{1}{16}) \Rightarrow \frac{1}{λ_{2}} = 1.097 \times 10^{7} (\frac{16 - 9}{144}) \Rightarrow \frac{1}{λ_{2}} = \frac{1.097 \times 10^{7} \times 7}{144} \Rightarrow λ_{2} = \frac{144}{1.097 \times 10^{7} \times 7} = 1875 nm$

Similarly, wavelength ( $λ$ ₃) for the transition from (n = 3) to (n = 2) is given by
When the transition is n₁ = 2 to n₂ = 3:
$\frac{1}{λ_{3}} = 1.097 \times 10^{7} (\frac{1}{4} - \frac{1}{9}) \Rightarrow \frac{1}{λ_{3}} = 1.097 \times 10^{7} (\frac{9 - 4}{36}) \Rightarrow \frac{1}{λ_{3}} = \frac{1.097 \times 10^{7} \times 5}{36} \Rightarrow λ_{3} = \frac{36 \times 10^{- 7}}{1.097 \times 5} = 656 nm$

Page No 384:

Answer:

Given:
Wavelength of photon, λ = 228 Å
Energy $(E)$ is given by
$E = \frac{h c}{λ}$
Here, c = Speed of light
h = Planck's constant

$∴ E = \frac{(6.63 \times 10^{- 34}) \times (3 \times 10^{8})}{(228 \times 10^{- 10})} = 0.0872 \times 10^{- 16} J$

As the transition takes place from n = 1 to n = 2, the excitation energy (E₁) will be
$E_{1} = R h c Z^{2} (\frac{1}{{n_{1}}^{2}} - \frac{1}{{n_{2}}^{2}}) E_{1} = (13.6 eV) \times Z^{2} \times (\frac{1}{1^{2}} - \frac{1}{2^{2}}) \Rightarrow E_{1} = (13.6 eV) \times Z^{2} \times \frac{3}{4}$
This excitation energy should be equal to the energy of the photon.
$∴ 13.6 \times \frac{3}{4} \times Z^{2} = 0.0872 \times 10^{- 16} Z^{2} = \frac{0.0872 \times 10^{- 16} \times 4}{13.6 \times 3 \times 1.6 \times 10^{- 19}} = 5.34 Z = \sqrt{5.34} = 2.3$

The ion may be helium.

Page No 384:

Answer:

Charge on the electron, q₁ = 1.6 $\times$ 10 ^{$- 19$}C
Charge on the nucleus, q₂ = 1.6 $\times$ 10 ^{$- 19$}C

Let r be the distance between the nucleus and the electron.
Coulomb force $(F)$ is given by
$F = \frac{q_{1} q_{2}}{4 π \in_{0} r^{2}}$ .....(1)
Here, q₁ = q₂ = q = 1.6 $\times$ 10 ^{$- 19$}C
Smallest distance between the nucleus and the first orbit, r = 0.53 $\times$ 10^{$-$ 10} m
$K = \frac{1}{4 {πε}_{0}}$ = 9 $\times$ 10⁹ Nm²C^{$-$ 2}
Substituting the respective values in (1), we get
$F = \frac{(9 \times 10^{9}) \times (1.6 \times 10^{- 19}) \times (1.6 \times 10^{- 19})}{{(0.53 \times 10^{- 10})}^{2}} = \frac{1.6 \times 1.6 \times 9 \times 10^{- 9}}{{(0.53)}^{2}} = 82.02 \times 10^{- 9} = 8.2 \times 10^{- 8} N$

Page No 384:

Answer:

(a)
The binding energy of hydrogen is given by
$E = \frac{13.6}{n^{2}} eV$
For binding energy of 0.85 eV,
${n_{2}}^{2} = \frac{13.6}{0.85} = 16 n_{2} = 4$
For binding energy of 10.2 eV,
${n_{1}}^{2} = \frac{13.6}{10.2} n_{1} = 1.15 \Rightarrow n_{1} = 2$
The quantum number of the upper and the lower energy state are 4 and 2, respectively.

(b) Wavelength of the emitted radiation $(λ)$ is given by

$\frac{1}{λ} = R (\frac{1}{{n_{1}}^{2}} - \frac{1}{{n_{2}}^{2}})$

Here,
R = Rydberg constant
n₁ and n₂ are quantum numbers.

$∴ \frac{1}{λ} = 1.097 \times 10^{7} (\frac{1}{4} - \frac{1}{16}) \Rightarrow λ = \frac{16}{1.097 \times 3 \times 10^{7}} = 4.8617 \times 10^{- 7} = 487 nm$

Page No 384:

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 $(λ)$ is given by
$\frac{1}{λ} = R (\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}})$
Here, R is the Rydberg constant, having the value of 1.097×10⁷ m^-1.

$\frac{1}{λ} = 1.097 \times 10^{7} [\frac{1}{{(1)}^{2}} - \frac{1}{{(2)}^{2}}] \Rightarrow \frac{1}{λ} = 1.097 \times 10^{7} [1 - \frac{1}{4}] \Rightarrow \frac{1}{λ} = 1.097 \times \frac{3}{4} \times 10^{7} \Rightarrow λ = \frac{4}{1.097 \times 3 \times 10^{7}} = 1.215 \times 10^{- 7} = 121.5 \times 10^{- 9} = 122 nm$

Page No 384:

Answer:

Energy (E) of the nth state of hydrogen atom is given by
E = $\frac{13.6}{n^{2}}$ eV

For n = 6,

$∴ E = \frac{- 13.6}{36} = - 0.377777777$ 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)
                                                        $= 13.6 - (0.37777 + 1.13) = 12.09 = 12.1 eV$

(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

$\frac{13.6}{n^{2}} = 1.507$

$∴ n = \sqrt{\frac{13.6}{1.507}} = \sqrt{9.024} \approx 3$

Page No 384:

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).
$E = 13.6 \times (\frac{1}{{n_{1}}^{2}} - \frac{1}{{n_{2}}^{2}}) eV$
Therefore, the excitation energy is given by

$E = 13.6 \times (\frac{1}{1^{2}} - \frac{1}{2^{2}}) eV E = 13.6 \times \frac{3}{4} eV = 10.2 eV$
Energy 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

Page No 384:

Answer:

Given:
Energy (E) of the ground state will be the energy acquired in the transition of the 2 excitation state to ground state.

$E_{1} = \frac{h c}{λ_{1}}$
Here,
h = Planck's constant
c = Speed of light
$λ_{1}$ = Wavelength of the radiation emitted when atoms come from the highest excited state to ground state

$∴ E_{1} = \frac{(6.63 \times 10^{- 34}) \times (3 \times 10^{8})}{(46 \times 10^{- 9})} J E_{1} = \frac{(6.63 \times 10^{- 34}) \times (3 \times 10^{8})}{(46 \times 10^{- 9}) \times (1.6 \times 10^{- 19})} eV = \frac{1242}{46} = 27 eV$

Energy in the first excitation state $(E_{2})$ will be the energy acquired in the transition of the highest energy state to the 2nd excitation state.

$E_{2} = \frac{h c}{λ_{n}}$

Here,
$λ_{n}$ = Wavelength of the radiation emitted when an atom comes from the highest energy state to the 2nd excitation state.
$E_{2} = \frac{h c}{λ_{n}} E_{2} = \frac{(6.63 \times 10^{- 34}) \times (3 \times 10^{8})}{(103.5 \times 10^{- 9})} J E_{2} = \frac{(6.63 \times 10^{- 34}) \times (3 \times 10^{8})}{(103.5 \times 10^{- 9}) \times (1.6 \times 10^{- 19})} eV = 12 eV$

Page No 384:

Answer:

(a) If the atom is excited to the principal quantum (n), then the number of transitions is given by
$\frac{n (n - 1)}{2}$
It is given that a total of 6 photons are emitted. Therefore, total number of transitions is 6.

∴ $\frac{n (n - 1)}{2}$ = 6
⇒ n = 4

Thus, the principal quantum number is 4 and the gas is in the 4th excited state.

Page No 384:

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.
$\Rightarrow m v r = \frac{n h}{2 π}$ (Here, n is an integer.)
$v = r ω \Rightarrow m r^{2} ω = \frac{n h}{2 π} \Rightarrow ω = \frac{n h}{2 π \times m \times r^{2}}$

$∴ ω = \frac{1 \times (6.63 \times 10^{- 34})}{2 \times (3.14) \times (9.1093 \times 10^{- 31}) \times {(0.53 \times 10^{- 10})}^{2}} = 0.413 \times 10^{17} rad / s = 4.13 \times 10^{16} rad / s$

Page No 384:

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 + ∆λ) < 8000.

Number of wavelengths in this range will be calculated in the following way:
$\frac{656.3 - 365}{8000} = 36$
Two lines will be extra for the first and last wavelength.
∴ Total number of lines = 36 + 2 = 38

Page No 384:

Answer:

Given:
Possible transitions:
From n₁ = 1 to n₂ = 3
n₁ = 2 to n₂ = 4

(a) Here, n₁ = 1 and n₂ = 3
Energy, $E = 13.6 (\frac{1}{{n_{1}}^{2}} - \frac{1}{{n_{2}}^{2}})$
$E = 13.6 (\frac{1}{1} - \frac{1}{9}) = 13.6 \times \frac{8}{9} . . . . . (1) Energy (E) is also given by E = \frac{h c}{λ}$

Here, h = Planck constant
c = Speed of the light
$λ$ = Wavelength of the radiation
$∴ E = \frac{6.63 \times 10^{- 34} \times 3 \times 10^{8}}{λ} . . . . . (2) Equating equation s (1) and (2), we have \Rightarrow λ = \frac{6.63 \times 10^{- 34} \times 3 \times 10^{8} \times 9}{13.6 \times 8} = 0.027 \times 10^{- 7} = 103 nm$

(b) Visible radiation comes in Balmer series.
As 'n' changes by 2, we consider n = 2 to n = 4.

Energy, $E_{1} = 13.6 (\frac{1}{{n_{1}}^{2}} - \frac{1}{{n_{2}}^{2}})$
$= 13.6 \times (\frac{1}{4} - \frac{1}{16}) = 2.55 eV$
If $λ_{1}$ is the wavelength of the radiation, when transition takes place between quantum number n = 2 to n = 4, then
255 = $\frac{1242}{λ_{1}}$
or λ₁ = 487 nm

Page No 384:

Answer:

Let v₀ be the velocity of the electron moving in the ground state and $r_{0}$ be the radius of the ground state.
Frequency of the revolution of electron in the circle is given by
$f = \frac{v_{0}}{2 π r_{0}}$

Frequency of the radiation emitted = Frequency of the revolution of electron
∴ Frequency of the radiation emitted = $\frac{v_{0}}{2 π r_{0}}$
Also, c = f $λ$
Here, c = Speed of light
$λ$ = Wavelength of the radiation emitted
$\Rightarrow$ $λ = \frac{c}{f}$
$∴ λ = \frac{2 π r_{0} c}{v_{0}} = \frac{2 \times (3.14) \times (53 \times 10^{- 12}) \times (3 \times 10^{8})}{(2.187 \times 10^{6})} = 45.686 \times 10^{- 12} m = 45.7 nm$

Page No 385:

Answer:

Average kinetic energy $(K)$ of the molecules in a gas at temperature $(T)$ is given by
K = $\frac{3}{2} k T$
Here,
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
$\Rightarrow$ 1.5 × 8.62 × 10⁻⁵ × T = 13.6
$\Rightarrow T = \frac{13.6}{1.5 \times 8.62 \times 10^{- 5}} = 1.05 \times 10^{5} K$
No, it is impossible for hydrogen to remain in molecular state at such a high temperature.

Page No 385:

Answer:

Given:
Wavelength of red light, $λ$ = 653.1 nm = 653.1 $\times$ 10^{$- 9$} m

Kinetic energy of H₂ molecules $(K)$ is given by
$K = \frac{3}{2} k T$    ...(1)
Here, k = 8.62 × 10⁻⁵ eV/K
      T = Temperature of H₂ molecules
Energy released $(E)$ when atom goes from ground state to n = 3 is given by
$E = 13.6 (\frac{1}{{n_{1}}^{2}} - \frac{1}{{n_{2}}^{2}})$
For ground state, n₁ = 1
Also, n₂ = 3
$∴ E = 13.6 (\frac{1}{1} - \frac{1}{9}) = 13.6 (\frac{8}{9}) . . . (2)$

Kinetic energy of H₂ molecules = Energy released when hydrogen atom goes from ground state to n = 3 state
$∴ \frac{3}{2} \times 8.62 \times 10^{- 5} \times T = \frac{13.6 \times 8}{9} \Rightarrow T = \frac{13.6 \times 8 \times 2}{9 \times 3 \times 8.62 \times 10^{- 5}} = 9.4 \times 10^{4} K$

Page No 385:

Answer:

Frequency of electron (f) is given by
$f = \frac{m e^{4}}{4 {\in_{0}}^{2} n^{3} h^{3}}$
Time period is given by
$T = \frac{1}{f} T = \frac{4 \in_{0}^{2} n^{3} h^{3}}{m e^{4}}$
Here,
h = Planck's constant
m = Mass of the electron
e = Charge on the electron
$ε_{0}$ = Permittivity of free space
$∴ T = \frac{4 \times (8.85 \times 10^{- 12}) \times {(2)}^{3} \times {(6.63 \times 10^{- 34})}^{3}}{(9.10 \times 10^{- 31}) \times {(1.6 \times 10^{- 16})}^{4}} = 12247.735 \times 10^{- 19} s$
Average life time of hydrogen, t = 10^{$-$ 8} s
Number of revolutions is given by
$N = \frac{t}{T} \Rightarrow N = \frac{10^{- 8}}{12247.735 \times 10^{- 19}}$
N = 8.2 $\times$ 10⁵revolution

Page No 385:

Answer:

Mass of the electron, m = 9.1×10^{$-$ 31}kg
Radius of the ground state, r = 0.53×10^{$-$ 10}m
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
$= e \times \frac{m e^{4}}{4 \in_{0}^{2} h^{3} n^{3}} \times (π r^{2} n^{2}) = \frac{m e^{5} \times (π r^{2} n^{2})}{4 \in_{0}^{2} h^{3} n^{3}}$
Here,
h = Planck's constant
e = Charge on the electron
$ε_{0}$ = Permittivity of free space
n = Principal quantum number

$∴ μ = \frac{(9.1 \times 10^{- 31}) {(1.6 \times 10^{- 19})}^{5} \times π \times {(0.53 \times 10^{- 10})}^{2}}{4 \times {(8.85 \times 10^{- 12})}^{2} \times {(6.64 \times 10^{- 34})}^{3} \times {(1)}^{3}} = 0.000917 \times 10^{- 20} = 9.176 \times 10^{- 24} A - m^{2}$

Page No 385:

Answer:

Mass of the electron, m = 9.1×10^{$-$ 31}kg
Radius of the ground state, r = 0.53×10^{$-$ 10}m
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
$= e \times \frac{m e^{4}}{4 \in_{0}^{2} h^{3} n^{3}} \times (π r_{0}^{2} n^{2}) = \frac{m e^{5} \times (π {r_{0}}^{2} n^{2})}{4 \in_{0}^{2} h^{3} n^{3}}$
Here,
h = Planck's constant
e = Charge on the electron
$ε_{0}$ = Permittivity of free space
n = Principal quantum number

Angular momentum of the electron in the hydrogen like atoms and ions (L) is given by
$L = m v r = \frac{n h}{2 π}$
Ratio of the dipole moment and the angular momentum is given by
$\frac{μ}{L} = \frac{e^{5} \times m \times π r^{2} n^{2}}{4 \in_{0} h^{3} n^{3}} \times \frac{2 π}{n h}$
$\frac{μ}{L} = \frac{{(1.6 \times 10^{- 19})}^{5} \times (9.10 \times 10^{- 31}) \times {(3.14)}^{2} \times {(0.53 \times 10^{- 10})}^{2}}{2 \times {(8.85 \times 10^{- 12})}^{2} \times {(6.63 \times 10^{- 34})}^{3} \times 1^{2}} \frac{μ}{L} = 3.73 \times 10^{10} C / kg$

Ratio of the magnetic dipole moment and the angular momentum do not depends on the atomic number 'Z'.
Hence, it is a universal constant.

Page No 385:

Answer:

Given:
Minimum wavelength of the light component present in the beam, $λ_{1}$ = 450 nm
Energy associated $(E_{1})$ with wavelength $(λ_{1})$ is given by
E₁ = $\frac{h c}{λ_{1}}$
Here,
c = Speed of light
h = Planck's constant
$∴ E_{1} = \frac{1242}{450}$
= 2.76 eV

Maximum wavelength of the light component present in the beam, $λ_{2}$ = 550 nm

Energy associated $(E_{2})$ with wavelength $(λ_{2})$ is given by
E₂ = $\frac{h c}{λ_{2}}$
$∴ E_{2} = \frac{1242}{550} = 2.228 = 2.26 eV$
The given range of wavelengths lies in the visible range.
∴ n₁ = 2, n₂ = 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.
$E'_{2} - E'_{3} = 13.6 (\frac{1}{4} - \frac{1}{9}) = \frac{12.6 \times 5}{30} = 1.9 eV E'_{2} - E'_{4} = 13.6 (\frac{1}{4} - \frac{1}{16}) = 2.55 eV E'_{2} - E'_{5} = 13.6 (\frac{1}{4} - \frac{1}{25}) = \frac{10.5 \times 21}{100} = 2.856 eV$
Only, E'₂ − E'₄ comes in the range of the energy provided. So the wavelength of light having 2.55 eV will be absorbed.
$λ = \frac{1242}{2.55} = 487.05 nm = 487 nm$
The wavelength 487 nm will be absorbed by hydrogen gas. So, wavelength 487 nm will have less intensity in the transmitted beam.

Page No 385:

Answer:

Energy of radiation (E) from the hydrogen atom is given by
$E = 13.6 (\frac{1}{{n_{1}}^{2}} - \frac{1}{{n_{2}}^{2}})$
Hydrogen atoms go through transition, n = 1 to n = 2.
The energy released is given by
$E = 13.6 (\frac{1}{1} - \frac{1}{4}) = 13.6 \times \frac{3}{4} = 10.2 eV$
For He,
Atomic no, Z = 2
Let us check the energy required for the transition in helium ions from n = 1 to n = 2.

∴ n₁ = 1 to n₂ = 2

Energy $(E_{1})$ of this transition is given by
$E_{1} = Z^{2} 13.6 (\frac{1}{{n_{1}}^{2}} - \frac{1}{{n_{2}}^{2}}) = 4 \times 13.6 (1 - \frac{1}{4}) = 40.8 eV$
E₁ > 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.
∴ n₁ = 1 to n₂ = 3

Energy $(E_{2})$ for this transition is given by
$E_{2} = Z^{2} \times 13.6 (\frac{1}{{n_{2}}^{2}} - \frac{1}{{n_{1}}^{2}}) = 4 \times 13.6 \times (\frac{1}{1} - \frac{1}{9}) = 48.3 eV$
It is clear that E₂ > E.
Hence, this transition of helium ions is not possible.

Similarly, transition from n₁ = 1 to n₂ = 4 is also not possible.

Let us check the energy required for the transition in helium ion from n = 2 to n = 3.
∴ n₁ = 2 to n₂ = 3
Energy $(E_{3})$ for this transition is given by
$E_{3} = 13.6 \times 4 (\frac{1}{4} - \frac{1}{9}) = \frac{20 \times 13.6}{36} = 7.56 eV$

Let us check the energy required for the transition in helium ion from n = 2 to n = 3.
∴ n₁= 2 to n₂ = 4
Energy $(E_{4})$ for this transition is given by
$E_{4} = 13.6 \times 4 (\frac{1}{4} - \frac{1}{16}) = 13.6 \times \frac{3}{4} = 10.2 eV$

We find that
E₃ < E
E₄ = E
Hence, possible transitions are from n = 2 to n = 3 and n = 2 to n = 4.

Page No 385:

Answer:

Given:
Wavelength of ultraviolet radiation, λ = 50 nm = 50 $\times$ 10^{$- 9$}m

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 n₁ = 1 to n₂ = ∞.

Work function, $W_{0} = 13.6 (\frac{1}{1} - \frac{1}{\infty}) = 13.6 eV$
Using Einstein's photoelectric equation, we get

$E = W_{0} + KE \Rightarrow \frac{h c}{λ} - 13.6 = KE (∵ E = \frac{h c}{λ}) \Rightarrow \frac{1242}{50} - 13.6 = KE \Rightarrow KE = 24.84 - 13.6 = 11.24 eV$

Page No 385:

Answer:

Given:
Wavelength of light, λ = 100 nm = 100 $\times$ 10^{$- 9$} m
Energy of the incident light $(E)$ is given by
$E = \frac{h c}{λ}$
Here,
h = Planck's constant
λ = Wavelength of light
$∴ E = \frac{1242}{100} E = 12.42 eV$

(a)
Let E₁ and E₂be the energies of the 1st and the 2nd state, respectively.
Let the transition take place from E₁ to E₂.
Energy absorbed during this transition is calculated as follows:
Here,
n₁ = 1
n₂ = 2
Energy absorbed (E') is given by
$E' = 13.6 (\frac{1}{{n_{1}}^{2}} - \frac{1}{{n_{2}}^{2}}) = 13.6 (\frac{1}{1} - \frac{1}{4}) = 13.6 \times \frac{3}{4} = 10.2 eV$
Energy left = 12.42 eV − 10.2 eV = 2.22 eV
Energy of the photon = $\frac{h c}{λ}$
Equating the energy left with that of the photon, we get
$2.22 eV = \frac{h c}{λ} 2.22 eV = \frac{1242}{λ}$
or λ = 559.45 = 560 nm
Let E₃be the energy of the 3rd state.
Energy absorbed for the transition from E₁to E₃ is given by
$E' = 13.6 (\frac{1}{{n_{1}}^{2}} - \frac{1}{{n_{2}}^{2}}) = 13.6 (\frac{1}{1} - \frac{1}{9}) = 13.6 \times \frac{8}{9} = 12.1 eV$
Energy absorbed in the transition from E₁to E₃ = 12.1 eV (Same as solved above)

Energy left = 12.42 − 12.1 = 0.32 eV

$0.32 = \frac{h c}{λ} = \frac{1242}{λ} λ = \frac{1242}{0.32} = 3881.2 = 3881 nm$
Let E₄ be the energy of the 4th state.
Energy absorbed in the transition from E₃to E₄ is given by
$E' = 13.6 (\frac{1}{{n_{1}}^{2}} - \frac{1}{{n_{2}}^{2}}) = 13.6 (\frac{1}{9} - \frac{1}{16}) = 13.6 \times \frac{7}{144} = 0.65 eV$
Energy 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
$11.77 = \frac{h c}{λ} or λ = \frac{1242}{11.77} = 105.52$
The 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:
$E = 10.2 eV \Rightarrow 10.2 = \frac{h c}{λ} or λ = \frac{1242}{10.2} = 121.76 nm \approx 121 nm E = 12.1 eV \Rightarrow 12.1 = \frac{h c}{λ} or λ = \frac{1242}{12.1} = 102.64 nm \approx 103 nm E = 0.65 eV \Rightarrow 0.65 = \frac{h c}{λ} or λ = \frac{1242}{0.65} = 1910.76 nm 1911 nm$
Thus, the wavelengths of the radiations detected are 103 nm, 121 nm and 1911 nm.

Page No 385:

Answer:

Given:

Work function of cesium surface, ϕ = 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,
$\frac{hc}{λ} = E + ϕ$
Here,
h = Planck's constant
c = Speed of light
$λ$ = Wavelength of light
$∴ \frac{hc}{λ} - 1.9 = 13.6 \Rightarrow \frac{1240}{λ} = 15.5 \Rightarrow λ = \frac{1240}{15.5} \Rightarrow λ = 80 nm$

(b) When the electron is excited from the states n₁ = 1 to n₂ = 2, energy absorbed $(E_{1})$ is given by
$E_{1} = 13.6 (\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}) E_{1} = 13.6 (1 - \frac{1}{4}) E_{1} = \frac{13.66 \times 3}{4}$

For Einstein's photoelectric equation,

$∴ \frac{hc}{λ} - 1.9 = \frac{13.6 \times 3}{4} \Rightarrow \frac{hc}{λ} = \frac{13.6 \times 3}{4} + 1.9 \Rightarrow \frac{1240}{λ} = 10.2 + 1.9 = 12.1 \Rightarrow λ = \frac{1240}{12.1} \Rightarrow λ = 102.47 = 102 nm$

(c) Excited atom will emit visible light if an electron jumps from the second orbitto third orbit, i.e. from n₁ = 2 to n₂ = 3. This is because Balmer series lies in the visible region.
Energy (E₂) of this transition is given by
$E_{2} = 13.6 (\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}) E_{2} = 13.6 (\frac{1}{4} - \frac{1}{9}) E_{2} = \frac{13.66 \times 5}{36}$

For Einstein's photoelectric equation,
$\frac{hc}{λ} - 1.9 = \frac{13.6 \times 5}{36} \Rightarrow \frac{hc}{λ} = \frac{13.6 \times 5}{36} + 1.9 \Rightarrow \frac{1240}{λ} = 1.88 + 1.9 = 3.78 \Rightarrow λ = \frac{1240}{3.78} \Rightarrow λ = 328.04 nm$

Page No 385:

Answer:

Given:
Distance travelled by the electron, d = 1.0 m
Wavelength of red light, $λ$ = 656.3 nm = 656.3 $\times$ 10^{$- 9$} m

Since the given wavelength lies in Balmer series, the transition that requires minimum energy is from n₁ = 3 to n₂ = 2.
Energy of this transition will be equal to the energy $(E_{1})$ that will be required for the transition from the ground state to n = 3.
$E_{1} = 13.6 (\frac{1}{{n_{1}}^{2}} - \frac{1}{{n_{2}}^{2}})$
$\Rightarrow E_{1} = 13.6 (1 - \frac{1}{9}) = \frac{13.6 \times 8}{9} = 12.09 eV$
Energy, E (eV) = 12.09 eV
∴ V = 12.09 V
Electric field, E = $\frac{V}{d}$ = $\frac{12.09}{1}$ = 12.09 V/m
∴ Minimum value of the electric field = 12.09 V/m = 12.1 V/m

Page No 385:

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.

Page No 385:

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 v₁ and v₂ 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 = mv₁ + mυ₂   ...(1)

Applying the conservation of mechanical energy, we get
$\frac{1}{2} m υ^{2} = \frac{1}{2} m υ_{1}^{2} + \frac{1}{2} m υ_{2}^{2} + ∆ E . . . (2)$

Using equation (1), we get
v² = (v₁ + v₂)²
v² $= υ_{1}^{2} + υ_{2}^{2} + 2 υ_{1} υ_{2}$      ...(3)

In equation(2), on multiplying both the sides by 2 and dividing both the sides by m.
$υ^{2} = υ_{1}^{2} + υ_{1}^{2} + 2 ∆ E / m . . . . (4)$

On comparing (4) and (3), we get
$[∵ 2 υ_{1} υ_{2} = \frac{2 ∆ E}{m}]$
(υ₁ − υ₂)² = (υ₁ + υ₂)² − 4υ₁υ₂
_{${(v_{1} - v_{2})}^{2} = v^{2} - \frac{4 ∆ E}{m}$}

For minimum value of υ,
v₁= v₂=0
Also, $v^{2} - \frac{4 ∆ E}{m} = 0$
$∴ v^{2} = \frac{4 ∆ E}{m} = \frac{4 \times 13.6 \times 1.6 \times 10^{- 19}}{1.67 \times 10^{- 27}} v = \sqrt{\frac{4 \times 13.6 \times 1.6 \times 10^{- 19}}{1.67 \times 10^{- 27}}} = 10^{4} \sqrt{\frac{4 \times 13.6 \times 1.6}{1.67}} = 7.2 \times 10^{4} m / s$

Page No 385:

Answer:

Given:
Mass of neutron, m = 1.67 × 10⁻²⁷ kg
Since neutron is moving with velocity $(v)$ , its energy $(E)$ is given by
$E = \frac{1}{2} m υ^{2}$
Let the energy absorbed be ∆E.
The condition for inelastic collision is given below:
$\frac{1}{2} m υ^{2} > 2 ∆ E$
$\Rightarrow ∆ E < \frac{1}{4} m υ^{2}$

Since 10.2 eV, energy is required for the first excited state.
∴ ∆E < 10.2 eV
$∴ 10.2 eV < \frac{1}{4} m v^{2}$
Thus, minimum speed of the neutron is given by
$\Rightarrow υ_{\min} = \sqrt{\frac{4 \times 10.2}{m}} eV \Rightarrow υ_{\min} = \sqrt{\frac{10.2 \times 1.6 \times 10^{19} \times 4}{1.67 \times 10^{- 27}}} = 6 \times 10^{4} m / \sec$

Page No 385:

Answer:

Given:
Wavelength of light emitted by hydrogen, λ = 656.3 nm
Mass of hydrogen atom, m = 1.67 × 10⁻²⁷ kg

(a) Momentum $(p)$ is given by
$P = \frac{h}{λ}$
Here,
h = Planck's constant
λ = Wavelength of light

$∴ p = \frac{6.63 \times 10^{- 34}}{656.3 \times 10^{- 9}} p = 0.01 \times 10^{- 25} p = 1 \times 10^{- 27} kgm / s$

(b) Momentum, p = mv
    Here,
m = Mass of hydrogen atom
v = Speed of atom
$∴$ 1 × 10⁻²⁷ = (1.67 × 10⁻²⁷)× υ
    $\Rightarrow υ = \frac{1}{1.67} = 0.598 = 0.6 m / s$

(c) Kinetic energy $(K)$ of the recoil of the atom is given by
      $K = \frac{1}{2} m v^{2}$
     Here,
m = Mass of the atom
v = Velocity of the atom
     $∴ K = \frac{1}{2} \times (1.67 \times 10^{- 27}) \times {(0.6)}^{2} J$
        $K = \frac{0.3006 \times 10^{- 27}}{1.6 \times 10^{- 19}} eV K = 1.9 \times 10^{- 9} eV$

Page No 385:

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,
$E = \frac{h c}{λ} \Rightarrow λ = \frac{h c}{E} . . . (i)$

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.
$E = \frac{h c}{λ'} + E_{R} \Rightarrow λ' = \frac{h c}{E - E_{R}} . . . (ii)$
Fractional change in the wavelength is given as,
$\frac{∆ λ}{λ} = \frac{λ' - λ}{λ} \Rightarrow \frac{∆ λ}{λ} = \frac{1}{λ} (\frac{h c}{E - E_{R}} - \frac{h c}{E}) \Rightarrow \frac{∆ λ}{λ} = \frac{E}{h c} (\frac{h c E_{R}}{E (E - E_{R})}) (∵ λ = \frac{h c}{E}) \Rightarrow \frac{∆ λ}{λ} = (\frac{E_{R}}{(E - E_{R}})$

Page No 385:

Answer:

The H_αlight 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
$E = 13.6 (\frac{1}{{n_{1}}^{2}} - \frac{1}{{n_{2}}^{2}})$ eV
Here, n₁ = 2, n₂ = 3
$∴ E = 13.6 \times (\frac{1}{4} - \frac{1}{9}) = \frac{13.6 \times 5}{36} = 1.89 eV = 1.90 eV$
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.

Page No 385:

Answer:

Let the maximum work function of the metal be W.

The energy liberated in the Balmer Series (E) is given by
$E = 13.6 (\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}})$

For maximum work function, maximum energy of Balmer's series is taken.

Now, n₁ = 2, n₁ = ∞

$∴ E = 13.6 (\frac{1}{2^{2}}) = 13.6 \times \frac{1}{4} = 3.4 eV$
Here,
W = E
Thus, maximum work function of metal is 3.4 eV.

Page No 385:

Answer:

Given:
Work function of cesium, $ϕ$ = 1.9 eV
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 eV

Page No 385:

Answer:

Wavelength of radiation coming from filter, λ = 440 nm
Work function of metal, ϕ = 2 eV
Charge of the electron, e = 1.6 $\times$ 10^{$- 9$}C
Let V₀ be the stopping potential.
From Einstein's photoelectric equation,
$\frac{h c}{λ} - ϕ = e V_{0}$
Here,
h =Planck constant
c = Speed of light
$λ$ = Wavelength of radiation

$\frac{4.14 \times 10^{- 15} \times 3 \times 10^{8}}{440 \times 10^{- 9}} - 2 eV = {eV}_{0} \Rightarrow e V_{0} = (\frac{1242}{440} - 2) eV = 0.823 eV . \Rightarrow V_{0} = 0.823 Volts$

Page No 385:

Answer:

Given:
Mass of the earth, m_e = 6.0 × 10²⁴ kg
Mass of the sun, m_s = 2.0 × 10³⁰ kg
Distance between the earth and the sun, d = 1.5 × 11¹¹ m

According to the Bohr's quantization rule,

Angular momentum, L = $\frac{n h}{2 π}$
$\Rightarrow m v r = \frac{n h}{2 π}$    ....(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
$m_{e}^{2} v^{2} r^{2} = \frac{n^{2} h^{2}}{4 π^{2}} . . . . (2)$
Gravitational force of attraction between the earth and the sun acts as the centripetal force.
$F = \frac{G m_{e} m_{s}}{r^{2}} = \frac{m_{e} v^{2}}{r}$
$\Rightarrow v^{2} = \frac{G m_{s}}{r}$      ....(3)
Dividing (2) by (3), we get
$m_{e}^{2} r = \frac{n^{2} h^{2}}{4 π^{2} G m_{s}}$

(a) For n = 1,

$r = \sqrt{\frac{h^{2}}{4 π^{2} G m_{s} m_{e}^{2}}} r = \sqrt{\frac{{(6.63 \times 10^{- 34})}^{2}}{4 \times {(3.14)}^{2} \times (6.67 \times 10^{- 11}) \times {(6 \times 10^{24})}^{2} \times (2 \times 10^{30})}} r = 2.29 \times 10^{- 138} m r = 2.3 \times 10^{- 138} m$

(b)
From (2), the value of the principal quantum number (n) is given by
$n^{2} = \frac{m_{e}^{2} \times r \times 4 \times π \times G \times m_{s}}{h^{2}} \Rightarrow n = \sqrt{\frac{m_{e}^{2} \times r \times 4 \times π \times G \times m_{s}}{h^{2}}}$
$n = \sqrt{\frac{{(6 \times 10^{24})}^{2} \times (1.5 \times 10^{11}) \times 4 \times {(3.14)}^{2} \times (6.67 \times 10^{- 11}) \times (2 \times 10^{30})}{{(6.6 \times 10^{- 34})}^{2}}} n = 2.5 \times 10^{74}$

Page No 386:

Answer:

According to Bohr's quantization rule,

Angular momentum of electron, L = $\frac{n h}{2 π}$
$\Rightarrow m v_{e} r = \frac{n h}{2 π} \Rightarrow v_{e} = \frac{n h}{2 π r m_{e}}$ ...(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,
$\frac{G m_{n} m_{e}}{r^{2}} = \frac{m_{e} v^{2}}{r} \Rightarrow \frac{G m_{n}}{r} = v^{2} . . . (2)$
Squaring (1) and dividing it by (2), we have
$\frac{m_{e}^{2} v^{2} r^{2}}{v^{2}} = \frac{n^{2} h^{2} r}{4 π^{2} G m_{n}} \Rightarrow m_{e}^{2} r^{2} = \frac{n^{2} h^{2} r}{4 π^{2} G m_{n}} \Rightarrow r = \frac{n^{2} h^{2}}{4 π^{2} G m_{n} m_{e}^{2}} v_{e} = \frac{n h}{2 π r m_{e}} \Rightarrow v_{e} = \frac{n h}{2 π m_{e} \times (\frac{n^{2} h^{2}}{4 π^{2} G m_{n} m_{e}^{2}})} \Rightarrow v_{e} = \frac{2 π G m_{n} m_{e}}{n h}$

$Kinetic e nergy of the electron, K = \frac{1}{2} m_{e} {v_{e}}^{2} = \frac{1}{2} m_{e} {(\frac{2 πG m_{n} m_{e}}{n h})}^{2} = \frac{4 π^{2} G^{2} m_{n}^{2} m_{e}^{3}}{2 n^{2} h^{2}}$

Potential energy of the neutron, $P = \frac{- G m_{n} m_{e}}{r}$

Substituting the value of r in the above expression,
$P = \frac{- G m_{e} m_{n} 4 π^{2} G m_{n} m_{e}^{2}}{n^{2} h^{2}} P = \frac{- 4 π^{2} G^{2} m_{n}^{2} m_{e}^{3}}{n^{2} h^{2}}$
Total energy = K + P $= - \frac{2 π^{2} G^{2} m_{n}^{2} m_{e}^{2}}{2 n^{2} h^{2}} = - \frac{π^{2} G^{2} m_{n}^{2} m_{e}^{2}}{n^{2} h^{2}}$

Page No 386:

Answer:

According to Bohr's quantization rule,
$m v r = \frac{n h}{2 π}$
'r' is minimum when 'n' has minimum value, i.e. 1.
$m v = \frac{n h}{2 π r} . . . (1) Again, r = \frac{m v}{q B} \Rightarrow m v = r q B . . . (2)$

From (1) and (2), we get
$r q B = \frac{n h}{2 π r} [From (1)] \Rightarrow r^{2} = \frac{n h}{2 π e B} [∴ q = e] \Rightarrow r = \sqrt{\frac{h}{2 πeB}} [n = 1]$

(b) For the radius of n^th orbit,
$r = \sqrt{\frac{n h}{2 π e B}}$

(c) $m v r = \frac{n h}{2 π}, r = \frac{m v}{q B}$
Substituting the value of 'r' in (1), we get
$m v \times \frac{m v}{q B} = \frac{n h}{2 π} \Rightarrow m^{2} v^{2} = \frac{h e B}{2 π} [n = 1, q = e] \Rightarrow v^{2} = \frac{h e B}{2 π m^{2}} v = \sqrt{\frac{h e B}{2 π m^{2}}}$

Page No 386:

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 n₁ = 2 and n₂ = 4.

We have the longest possible wavelength for minimum energy.
Energy of the light emitted (E) is given by
$E = 13.6 (\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}) E = 13.6 [\frac{1}{{(2)}^{2}} - \frac{1}{{(4)}^{2}}] E = 13.6 (\frac{1}{4} - \frac{1}{16}) E = \frac{13.6 \times 12}{64} = 2.55 eV$
Equating the calculated energy with that of photon, we get
$2.55 eV = \frac{h c}{λ} λ = \frac{h c}{2.55} = \frac{1242}{2.55} nm = 487.05 nm = 487 nm$

Page No 386:

Answer:

Let the frequency emitted by the atom at rest be ν₀.

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
$Frequency of the emitted radiation, ν = ν_{0} (\frac{1 + \frac{u}{c}}{1 - \frac{u}{c}}) Since u < < < c, ν = ν_{0} (\frac{1 + \frac{u}{c}}{1}) ν = ν_{0} (1 + \frac{u}{c})$
Ratio of frequencies of the emitted radiation, $\frac{ν}{ν_{0}} = (1 + \frac{u}{c})$

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