5 Marks Questions - Physics STD 12 Science Questions - Vidyadip

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5 Marks Questions

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17 questions · self-marked practice — reveal the answer and mark yourself.

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 $ν_0$ 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.
$\text{n} = 1 \text{ to n} = 2$
$\text{E}_1=4\times13.6\Big(1-\frac{1}{4}\Big)=40.8\text{eV}$ $[E_1 > E$ hence it is not possible$]$
$\text{n} = 1 \text{ to n} = 3$
$\text{E}_2=4\times13.6\Big(1-\frac{1}{9}\Big)=48.3\text{eV}$ $[E_2 > E$ 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, $E_3 < E$ and $E_4 = E$
Hence $E_3$ and $E_4$ can be possible.

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Question 35 Marks

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.

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Question 45 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

$\phi=1.9\text{eV}$

The hydrogen is ionized,

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

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

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

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

For the electron to be excited from,

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

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

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

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

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

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

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

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 55 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 $n^{th}$ 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 65 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 $= 1.67 \times 10^{-27}kg.$

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 75 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 $n_1$ to $n_2$ 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 85 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 95 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 $= 6.0 \times 10^{-24}kg$. Mass of the sun $= 2.0 \times 10^{30}kg, $ earth-sun distance $= 1.5 \times 10^{11}m.$

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

For n = 1

$\text{r}=\sqrt{\frac{\text{h}^2}{4\pi^2\text{GMsMe}^2}}=2.29\times10^{-138}\text{m}$
$\text{r}=2.3\times10^{-138}\text{m}$

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

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

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Question 105 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 115 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,
$\text{n}_1=1,\text{ n}_2=\infty$

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

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

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

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

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

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

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

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

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

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Question 125 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 $n_2 = 3$) 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 $= 1097 \times 107m^{-1}$
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 $n_2 = 2$) 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 135 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

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

The possible transitions may be $E_1$ to $E_2$

$E_1$ to $E_2$, energy absorbed = 10.2eV
Energy left $= 12.42 - 10.2 = 2.22eV$
$2.22\text{eV}=\frac{\text{hc}}{\lambda}=\frac{1242}{\lambda}$ or $\lambda=559.45=560\text{nm}$
$E_1$ to $E_3$, Energy absorbed $= 12.1eV$
Energy left $= 12.42 - 12.1 = 0.32eV$
$0.32=\frac{\text{hc}}{\lambda}=\frac{1242}{\lambda}$ or $\lambda=\frac{1242}{0.32}=3881.2=3881\text{nm}$
$E_3$ to $E_4$, Energy absorbed = 0.65
Energy left $= 12.42 - 0.65 = 11.77eV$
$11.77=\frac{\text{hc}}{\lambda}=\frac{1242}{\lambda}$ or $\lambda=\frac{1242}{11.77}=105.52$

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 145 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 $= 1.097 \times 10^7m^{-1}$
$n_1 =$ Quantum number of final state
$n_2 =$ Quantum number of initial state

For transition from n = 3 to n = 2,

Here, $n_1 = 2, n_2 = 3$
$\frac{1}{\lambda}=1.09737\times10^{7}\times\Big(\frac{1}{4}-\frac{1}{9}\Big)$
$\lambda=\frac{36}{5\ \times\ 1.09737\ \times\ 10^{7}}$
$=6.56\times10^{-7}=656\text{nm}$

For transition from n = 5 to n = 4,

Here, $n_1 = 4, n_2 = 5$
$\frac{1}{\lambda}=1.09737\times10^{-7}\Big(\frac{1}{16}-\frac{1}{25}\Big)$
$\lambda=\frac{400}{1.09737\ \times10^7\ \times\ 9}$
$\lambda=4050\text{nm}$

For transition from n = 10 to n = 9,

Here, $n_1 = 9, n_2 = 10$
$\frac{1}{\lambda}=1.09737\times10^7\Big(\frac{1}{81}-\frac{1}{100}\Big)$
$\lambda=\frac{81\ \times\ 100}{19\ \times\ 1.09737\ \times\ 10^7}$
$\lambda=38849\text{nm}$

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Question 155 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 165 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, $\text{r}=\frac{\text{mv}}{\text{qB}},\text{ or }\text{mv}=\text{rqB}\ ...(\text{ii})$

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]

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

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 175 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 $= 1.67 \times 10^{-27}kg.$

Answer

The hydrogen atoms after collision move with speeds $v_1$ and $v_2$
$\text{mv}=\text{mv}_1+\text{mv}_2\ ....(\text{i})$
$\frac{1}{2}\text{mv}^2=\frac{1}{2}\text{mv}^2_1+\frac{1}{2}\text{mv}^2_2+\Delta\text{E}\ ....(\text{ii})$
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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