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Physics 5 Marks Questions

Bohr’s Model and Physics of Atom · Gujarat Board (GSEB) STD 12 Science (GSEB - English Medium). 17 questions.

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Question 15 Marks
Consider an excited hydrogen atom in state n moving with a velocity u(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 ν0 emitted if the atom were at rest.
Answer
Velocity of hydrogen atom in state ‘n’ = u
Also the velocity of photon = u
But u << C
Here the photon is emitted as a wave.
So its velocity is same as that of hydrogen atom i.e. u.
$\therefore$ Accounding to Doppler’s effect,
Frequency $\text{v}=\text{v}_0\bigg(\frac{1+\frac{\text{u}}{\text{c}}}{1-\frac{\text{u}}{\text{c}}}\bigg)$
as u <<< C
$1-\frac{\text{u}}{\text{c}}=\text{q}$
$\therefore\ \text{v}=\text{v}_0\bigg(\frac{1+\frac{\text{u}}{\text{c}}}{1}\bigg)=\text{v}_0\Big(1+\frac{\text{u}}{\text{c}}\Big)$
$\text{v}=\text{v}_0\Big(1+\frac{\text{u}}{\text{c}}\Big)$
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Question 25 Marks
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
From transitions n = 2 to n = 1
$\text{E}=13.6\Big(\frac{1}{1}-\frac{1}{4}\Big)=13.6\times\frac{3}{4}=10.2\text{eV}$
Let in check the transitions possible on He.
$\text{n} = 1 \text{ to n} = 2$
$\text{E}_1=4\times13.6\Big(1-\frac{1}{4}\Big)=40.8\text{eV}$ [E1 > E hence it is not possible]
$\text{n} = 1 \text{ to n} = 3$
$\text{E}_2=4\times13.6\Big(1-\frac{1}{9}\Big)=48.3\text{eV}$ [E2 > E hence impossible]
Similarly n = 1 to n = 4 is also not possible.
$\text{n} = 2 \text{ to n} = 3$
$\text{E}_3=4\times13.6\Big(\frac{1}{4}-\frac{1}{9}\Big)=7.56\text{eV}$
$\text{n} = 2 \text{ to n} = 4$
$\text{E}_4=4\times13.6\Big(\frac{1}{4}-\frac{1}{16}\Big)10.2\text{eV}$
As, E3 < E and E4 = E
Hence E3 and E4 can be possible.
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Question 35 Marks
A beam of monochromatic light of wavelength $\lambda$ ejects photoelectrons from a cesium surface $\big(\Phi=1.9\text{eV}\big).$ These photoelectrons are made to collide with hydrogen atoms in ground state. Find the maximum value of $\lambda$ 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
$\phi=1.9\text{eV}$

  1. The hydrogen is ionized,

$\text{n}_1=1,\text{n}_2=\infty$

Energy required for ionization $=13.6\bigg(\frac{1}{\text{n}^2_1}-\frac{1}{\text{n}^2_1}\bigg)=13.6$

$\frac{\text{hc}}{\lambda}-1.9=13.6$

$\lambda=80.1\text{nm}=80\text{nm}$

  1. For the electron to be excited from,

$\text{n}_1\text{ to n}_2=2$

$\text{E}=13.6\bigg(\frac{1}{\text{n}^2_1}-\frac{1}{\text{n}^2_1}\bigg)=13.6\Big(1-\frac{1}{4}\Big)=\frac{13.6\times3}{4}$

$\frac{\text{hc}}{\lambda}-1.9=\frac{13.6\times3}{4}$

$\lambda=\frac{1242}{12.1}=102.64=102\text{nm}$

  1. $\text{E}_2=13.6\bigg(\frac{1}{\text{n}^2_1}-\frac{1}{\text{n}^2_1}\bigg)$

$\text{E}_2=13.6\Big(\frac{1}{4}-\frac{1}{9}\Big)$

$\text{E}_2=\frac{13.66\times5}{36}$

For Einstein's photoelecrtic equation,

$\frac{\text{hc}}{\lambda}-1.9=\frac{13.6\times5}{36}$

$\frac{\text{hc}}{\lambda}=\frac{13.6\times5}{36}+1.9$

$\frac{1240}{\lambda1}=1.88+1.9=3.78$

$\lambda=\frac{1240}{3.78}$

$\lambda=328.04\text{nm}$

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Question 45 Marks
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 10eV when the electron is widely separated from the proton? Can we still write $\text{E}_\text{n}=\frac{\text{E}_1}{\text{n}^2}?\text{ r}_\text{n}=\text{a}_0\text{n}^2?$ 
Answer
Energy of nth state of hydrogen is given by,
$\text{E}_\text{n}=\frac{-13.6}{\text{n}^2}\text{eV}$
Energy of first excited state (n = 2) of hydrogen, $\text{E}_\text{n}=\frac{-13.6}{\text{n}^2}\text{eV}=-3.4\text{eV}$
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 10eV when the electron is widely separated from the proton so here our refrence point energy is 10eV. Earlier The energy of first excited state was -3.4eV 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,
$\text{E}_1=-3.4\text{eV}-10\text{eV}=-13.4\text{eV}$
We still write $\text{E}_\text{n}=\frac{\text{E}_1}{\text{n}^2},\text{ or }\text{r}_\text{n}=\text{a}_0\text{n}^2$ because these formulas are independent of the refrence point energy.
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Question 55 Marks
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 n1 to n2 is given by,
$\text{f}=\text{k}\bigg(\frac{1}{\text{n}^{2}_1}-\frac{1}{\text{n}^{2}_2}\bigg)$
Here, k is a constant.
For the series limit of Lyman series,
$\text{n}_1=1$
$\text{n}_2=\infty$
Frequency, $\text{f}_1=\text{k}\Big(\frac{1}{1^2}-\frac{1}{\infty}\Big)=\text{k}$
For the first line of Lyman series,
$\text{n}_1=1$
$\text{n}_2=2$
Frequency, $\text{f}_2=\text{k}\Big(\frac{1}{1^2}-\frac{1}{2^2}\Big)=\frac{3\text{k}}{4}$
For series limit of Balmer series,
$\text{n}_1=2$
$\text{n}_2=\infty$
$\text{f}_1=\text{k}\Big(\frac{1}{2^2}-\frac{1}{\infty}\Big)=\frac{\text{k}}{4}$
$\text{f}_1-\text{f}_3=\text{f}_2$
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.
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Question 65 Marks
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 neutron-electron system.
Answer
$\text{m}_\text{e}\text{Vr}=\frac{\text{nh}}{2\pi}\ ...(\text{i})$
$\frac{\text{GM}_\text{n}\text{M}_\text{e}}{\text{r}^2}=\frac{\text{m}_\text{e}\text{V}^2}{\text{r}}$
$\frac{\text{GM}_\text{n}}{\text{r}}=\text{v}^2\ ...(\text{ii})$
Squaring (ii) and dividing it with (i)
$\frac{\text{m}_\text{e}^2\text{v}^2\text{r}^2}{\text{v}^2}=\frac{\text{n}^2\text{h}^2\text{r}}{4\pi^2\text{Gm}_\text{n}}$
$\text{me}^2\text{r}=\frac{\text{n}^2\text{h}^2\text{r}}{4\pi^2\text{Gm}_\text{n}}$
$\text{r}=\frac{\text{n}^2\text{h}^2\text{r}}{4\pi^2\text{Gm}_\text{n}\text{me}^2}$
$\text{v}=\frac{\text{nh}}{2\pi\text{rm}_\text{e}}$ [from (i)]
$\text{v}=\frac{\text{nh4}\pi^2\text{GM}_\text{n}\text{M}^2_\text{e}}{2\pi\text{M}_\text{e}\text{n}^2\text{h}^2}=\frac{2\pi\text{GM}_\text{n}\text{M}_\text{e}}{\text{nh}}$
$\text{KE}=\frac{1}{2}\text{m}_\text{e}\text{V}^2$
$=\frac{1}{2}\text{m}_\text{e}\frac{\big(2\pi\text{GM}_\text{n}\text{M}_\text{e}\big)^2}{\text{nh}}=\frac{4\pi^2\text{G}^2\text{M}^2_\text{n}\text{M}^3_\text{e}}{2\text{n}^2\text{h}^2}$
$\text{PE}=\frac{-\text{GM}_\text{n}\text{M}_\text{e}}{\text{r}}$
$=\frac{-\text{GM}_\text{n}\text{M}_\text{e}4\pi^2\text{GM}_\text{n}\text{M}^2_\text{e}}{\text{n}^2\text{h}^2}=\frac{-4\pi^2\text{G}^2\text{M}^2_\text{n}\text{M}^3_\text{e}}{\text{n}^2\text{h}^2}$
Total energy $=\text{KE}+\text{PE}=\frac{2\pi^2\text{G}^2\text{M}^2_\text{n}\text{M}^3_\text{e}}{2\text{n}^2\text{h}^2}$
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Question 75 Marks
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 × 10-24kg. Mass of the sun = 2.0 × 1030kg, earth-sun distance = 1.5 × 1011m.
Answer
Mass of Earth = Me = 6.0 × 1024kg

Mass of Sun = Ms = 2.0 × 1030kg

Earth-Sun dist = 1.5 × 1011m

$\text{mvr}=\frac{\text{nh}}{2\pi}\text{ or m}^2\text{v}^2\text{r}^2=\frac{\text{n}^2\text{h}^2}{4\pi^2}\ ....(\text{i})$

$\frac{\text{GMeMs}}{\text{r}^2}=\frac{\text{mev}^2}{\text{r}}\text{ or v}^2=\frac{\text{GM}}{\text{s/r}}\ ....(\text{ii})$

Dividing (i) and (ii),

We get $\text{me}^2\text{r}=\frac{\text{n}^2\text{h}^2}{4\pi^2\text{GMs}}$

  1. For n = 1

$\text{r}=\sqrt{\frac{\text{h}^2}{4\pi^2\text{GMsMe}^2}}=2.29\times10^{-138}\text{m}$

$\text{r}=2.3\times10^{-138}\text{m}$

  1. $\text{n}^2=\frac{\text{Me}^2\times\text{r}\times4\times\pi^2\times\text{G}\times\text{Ms}}{\text{h}^2}$

$\text{n}^2=2.5\times10^{74}$

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Question 85 Marks
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,
$\text{R}=\frac{\text{A}}{\text{B}\rho(\nu)}=\text{e}\frac{\text{h}\nu}{\text{kT}}-1$
For microwave region
$\nu=10^{10}(\text{say})$
$\text{R}=\text{e}^{\frac{6.6\times10^{-34}\times10^{10}}{1.38\times10^{-23}\times300}}-1$
$\text{R}=\text{e}^{0.0016}-1$
$\text{R}=0.0016$
This implies that stimulated transition dominate in this region.
For visible region,
$\nu=10^{15}$
$\text{R}=\text{e}^{160}-1$
$\text{R}>>1$
So here spontaneous transition dominate.
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Question 95 Marks
Calculate the smallest wavelength of radiation that may be emitted by (a) hydrogen, (b) He+ and (c) Li++
Answer
Small wave length is emitted i.e. longest energy,

$\text{n}_1=1,\text{ n}_2=\infty$

  1. $\frac{1}{\lambda}=\text{R}\bigg(\frac{1}{\text{n}^2_1-\text{n}^2_2}\bigg)$

$\Rightarrow\frac{1}{\lambda}=1.1\times10^7\Big(\frac{1}{1}-\frac{1}{\infty}\Big)$

$\Rightarrow\lambda=\frac{1}{1.1\times10^7}=\frac{1}{1.1}\times10^{-7}$

$\Rightarrow\lambda=0.909\times10^{-7}$

$\Rightarrow\lambda=90.9\times10^{-8}=91\text{nm}$

  1. $\frac{1}{\lambda}=\text{z}^2\text{R}\bigg(\frac{1}{\text{n}^2_1}-\frac{1}{\text{n}^2_2}\bigg)$

$\lambda=\frac{1}{1.1\times10^{-7}\text{z}^2}=\frac{91\text{nm}}{4}=23\text{nm}$

  1. $\frac{1}{\lambda}=\text{z}^2\text{R}\bigg(\frac{1}{\text{n}^2_1}-\frac{1}{\text{n}^2_2}\bigg)$

$\Rightarrow\lambda=\frac{91\text{nm}}{\text{z}^2}=\frac{91}{9}=10\text{nm}$

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Question 105 Marks
How many wavelengths are emitted by atomic hydrogen in visible range (380nm-780nm)? In the range 50nm to 100nm?
Answer
Balmer series contains wavelengths ranging from 364nm (for n2 = 3) to $655\text{nm }(\text{n}_2)=\infty$ So, the given range of wavelength (380-780nm) lies in the Balmer series.
The wavelength in the Balmer series can be found by,
$\frac{1}{\lambda}=\text{R}\Big(\frac{1}{2^2}-\frac{1}{\text{n}^2}\Big)$
Here, R = Rydberg's constant = 1097 × 107m-1
The wavelength for the transition from n = 3 to n = 2 is given by,
$\frac{1}{\lambda_1}=\text{R}\Big(\frac{1}{2^2}-\frac{1}{\text{3}^2}\Big)$
$\lambda_1=656.3\text{nm}$
The wavelength for the transition from n = 4 to n = 2 is given by,
$\frac{1}{\lambda_2}=\text{R}\Big(\frac{1}{2^2}-\frac{1}{\text{4}^2}\Big)$
$\lambda_2=486.1\text{nm}$
The wavelength for the transition from n = 5 to n = 2 is given by,
$\frac{1}{\lambda_3}=\text{R}\Big(\frac{1}{2^2}-\frac{1}{\text{5}^2}\Big)$
$\lambda_3=434.0\text{nm}$
The wavelength for the transition from n = 6 to n = 2 is given by,
$\frac{1}{\lambda_4}=\text{R}\Big(\frac{1}{2^2}-\frac{1}{\text{6}^2}\Big)$
$\lambda_4=410.2\text{nm}$
The eavelength for the transition from n = 7 to n = 2 is given by,
$\frac{1}{\lambda_5}=\text{R}\Big(\frac{1}{2^2}-\frac{1}{\text{7}^2}\Big)$
$\lambda_5=397.0\text{nm}$
Thus, the wavelengths emitted by the atomic hydrogen in visible range (380-780 nm) are 5. Lyman series contains wavelengths ranging from 91nm (for n2 = 2) to $121\text{nm }(\text{n}_2)=\infty$ 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,
$\frac{1}{\lambda}=\text{R}\Big(\frac{1}{1^2}-\frac{1}{\text{n}^2}\Big)$
The wavelength for the transition from n = 2 to n = 1 is given by,
$\frac{1}{\lambda_1}=\text{R}\Big(\frac{1}{1^2}-\frac{1}{2^2}\Big)$
$\lambda_1=122\text{nm}$
The wavelength for the transition from n = 3 to n = 1 is given by,
$\frac{1}{\lambda_2}=\text{R}\Big(\frac{1}{1^2}-\frac{1}{3^2}\Big)$
$\lambda_2=103\text{nm}$
The wavelength for the transition from n = 4 ton = 1 is given by,
$\frac{1}{\lambda_3}=\text{R}\Big(\frac{1}{1^2}-\frac{1}{4^2}\Big)$
$\lambda_3=97.3\text{nm}$
The wevelength for the transion from n = 5 to n = 1 is given by,
$\frac{1}{\lambda_4}=\text{R}\Big(\frac{1}{1^2}-\frac{1}{5^2}\Big)$
$\lambda_4=95.0\text{nm}$
The eavelength for the transition from n = 6 to n = 1 is given by,
$\frac{1}{\lambda_5}=\text{R}\Big(\frac{1}{1^2}-\frac{1}{6^2}\Big)$
$\lambda_5=93.8\text{nm}$
So, it can be noted that the number of wevelengths laying between 50nm to 100nm are 3
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Question 115 Marks
A parallel beam of light of wavelength 100nm 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
$\lambda=100\text{nm}$

$\text{E}=\frac{\text{hc}}{\lambda}=\frac{1242}{100}=12.42\text{eV}$

  1. The possible transitions may be E1 to E2

E1 to E2, energy absorbed = 10.2eV

Energy left = 12.42 - 10.2 = 2.22eV

$2.22\text{eV}=\frac{\text{hc}}{\lambda}=\frac{1242}{\lambda}$ or $\lambda=559.45=560\text{nm}$

E1 to E3, Energy absorbed = 12.1eV

Energy left = 12.42 - 12.1 = 0.32eV

$0.32=\frac{\text{hc}}{\lambda}=\frac{1242}{\lambda}$ or $\lambda=\frac{1242}{0.32}=3881.2=3881\text{nm}$

E3 to E4, Energy absorbed = 0.65

Energy left = 12.42 - 0.65 = 11.77eV

$11.77=\frac{\text{hc}}{\lambda}=\frac{1242}{\lambda}$ or $\lambda=\frac{1242}{11.77}=105.52$

  1. The energy absorbed by the H atom is now radiated perpendicular to the incident beam.

$10.2=\frac{\text{hc}}{\lambda}\text{ or }\lambda=\frac{1242}{10.2}=121.76\text{nm}$

$12.1=\frac{\text{hc}}{\lambda}\text{ or }\lambda=\frac{1242}{12.1}=102.64\text{nm}$

$0.65=\frac{\text{hc}}{\lambda}\text{ or }\lambda=\frac{1242}{0.65}=1910.76\text{nm}$

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Question 125 Marks
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 $(\lambda)\lambda$ of the radiation is given by,

$\frac{1}{\lambda}=\text{R}\Big(\frac{1}{\text{n}_1^2}-\frac{1}{\text{n}^2_2}\Big)$

Here, R = Rydberg constant = 1.097 × 107m-1

n1 = Quantum number of final state

n2 = Quantum number of initial state

  1. For transition from n = 3 to n = 2,

Here, n1 = 2, n2 = 3

$\frac{1}{\lambda}=1.09737\times10^{7}\times\Big(\frac{1}{4}-\frac{1}{9}\Big)$

$\lambda=\frac{36}{5\ \times\ 1.09737\ \times\ 10^{7}}$

$=6.56\times10^{-7}=656\text{nm}$

  1. For transition from n = 5 to n = 4,

Here, n1 = 4, n2 = 5

$\frac{1}{\lambda}=1.09737\times10^{-7}\Big(\frac{1}{16}-\frac{1}{25}\Big)$

$\lambda=\frac{400}{1.09737\ \times10^7\ \times\ 9}$

$\lambda=4050\text{nm}$

  1. For transition from n = 10 to n = 9,

Here, n1 = 9, n2 = 10

$\frac{1}{\lambda}=1.09737\times10^7\Big(\frac{1}{81}-\frac{1}{100}\Big)$

$\lambda=\frac{81\ \times\ 100}{19\ \times\ 1.09737\ \times\ 10^7}$

$\lambda=38849\text{nm}$

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Question 135 Marks
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 × 10-27kg.
Answer
Energy of the neutron is $\frac{1}{2}\text{mv}^2$
The condition for inelastic collision is,
$\frac{1}{2}\text{mv}^2>2\Delta\text{E}$
$\Delta\text{E}=\frac{1}{2}\text{mv}^2$
$\Delta\text{E}$ is the energy absorbed.
Energy required for first excited state is 10.2ev.
$\therefore\ \Delta\text{E}<10.2\text{ev}$
$\therefore10.2\text{ev}<\frac{1}{2}\text{mv}^2$
$\text{V}_\text{min}=\sqrt{\frac{4\times10.2}{\text{m}}}\text{ev}$
$\therefore\text{v}=\sqrt{\frac{10.2\times1.6\times10^{-19}\times4}{1.67\times10^{-27}}}=6\times10^4\text{m/sec}$
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Question 145 Marks
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,

$\text{mvr}=\frac{\text{nh}}{2\pi}$

‘r’ is less when ‘n’ has least value i.e. 1

or, $\text{mv}=\frac{\text{nh}}{2\pi\text{R}}\ ...(\text{i})$

Again, $\text{r}=\frac{\text{mv}}{\text{qB}},\text{ or }\text{mv}=\text{rqB}\ ...(\text{ii})$

  1. From (i) and (ii)

$\text{rqB}=\frac{\text{nh}}{2\pi\text{r}}$ [q = e]

$\text{r}^2=\frac{\text{nh}}{2\pi\text{eB}}$

$\text{r}=\sqrt{\frac{\text{h}}{2\pi\text{eB}}}$ [here n = 1]

  1. For the radius of nth orbit, $\text{r}=\sqrt{\frac{\text{nh}}{2\pi\text{eB}}}$
  2. $\text{mvr}=\frac{\text{nh}}{2\pi},\ \text{r}=\frac{\text{mv}}{\text{qB}}$

Substituting the value of ‘r’ in (i)

$\text{mv}\times\frac{\text{mv}}{\text{qB}}=\frac{\text{nh}}{2\pi}$

$\text{m}^2\text{v}^2=\frac{\text{nheB}}{2\pi\text{m}^2}$ [n = 1, q = e]

$\text{v}^{2}=\frac{\text{heB}}{2\pi\text{m}^2}$

$\text{v}=\sqrt{\frac{\text{heB}}{2\pi\text{m}^2}}$

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Question 155 Marks
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 × 10-27kg.
Answer
The hydrogen atoms after collision move with speeds v1 and v2
$\text{mv}=\text{mv}_1+\text{mv}_2\ ....(\text{i})$
$\frac{1}{2}\text{mv}^2=\frac{1}{2}\text{mv}^2_1+\frac{1}{2}\text{mv}^2_2+\Delta\text{E}\ ....(\text{ii})$
From (i), $\text{v}^2=(\text{v}_1+\text{v}_2)^2=\text{v}^2_1+\text{v}^2_2+2\text{v}_1\text{v}_2$
From (ii), $\text{v}^2=\text{v}^2_1+\text{v}^2_2+\frac{2\Delta\text{E}}{\text{m}}$
$=2\text{v}_1\text{v}_2=\frac{2\Delta\text{E}}{\text{m}}\ ...(\text{iii})$
$(\text{v}_1-\text{v}_2)^2=\big(\text{v}_1+\text{v}_2\big)^2-4\text{v}_1\text{v}_2$
$(\text{v}_1-\text{v}_2)=\text{v}^2-\frac{4\Delta\text{E}}{\text{m}}$
For minimum value of ‘v’
$\text{v}_1=\text{v}_2$
$\text{v}^2-\Big(\frac{4\Delta\text{E}}{\text{m}}\Big)=0$
$\text{v}^2=\frac{4\Delta\text{E}}{\text{m}}=\frac{4\times13.6\times1.6\times10^{-19}}{1.67\times10^{-27}}$
$\text{v}=\sqrt{\frac{4\times13.6\times1.6\times10^{-19}}{1.67\times10^{-27}}}=7.2\times10^4\text{m/s}$
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Question 165 Marks
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
$\text{n}_1-4\xrightarrow{\ \ \ }\text{n}_2=2$
$\text{n}_1=4\xrightarrow{\ \ \ }3\xrightarrow{\ \ \ }2$
$\frac{1}{\lambda}=1.097\times10^7\Big(\frac{1}{16}-\frac{1}{4}\Big)$
$\Rightarrow\frac{1}{\lambda}=1.097\times10^7\Big(\frac{1-4}{16}\Big)$
$\Rightarrow\frac{1.097\times10^7\times3}{16}$
$\Rightarrow\lambda=\frac{16\times10^{-7}}{3\times1.097}=4.8617\times10^{-7}$
$\Rightarrow\lambda=1.861\times10^{-9}=487\text{nm}$
$\text{n}_1 =4\text{ and }\text{n}_2=3$
$\frac{1}{\lambda}=1.097\times10^7\Big(\frac{1}{16}-\frac{1}{9}\Big)$
$\Rightarrow\frac{1}{\lambda}=1.097\times10^7\Big(\frac{9-16}{144}\Big)$
$\Rightarrow\frac{1.097\times10^7\times7}{144}$
$\Rightarrow\lambda=\frac{144}{7\times1.097\times10^7}=1875\text{nm}$
$\text{n}_1=3\xrightarrow{\ \ \ }\text{n}_2=2$
$\frac{1}{\lambda}=1.097\times10^7\Big(\frac{1}{9}-\frac{1}{4}\Big)$
$\Rightarrow\frac{1}{\lambda}=1.097\times10^7\Big(\frac{4-9}{36}\Big)$
$\Rightarrow\frac{1.097\times10^7\times5}{66}$
$\Rightarrow\lambda=\frac{36\times10^{-7}}{5\times1.097}=656\text{nm}$
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Question 175 Marks
We have stimulated emission and spontaneous emission. Do we also have stimulated absorption and spontaneous absorption?
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.

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