• MCQ (Single correct option)

• MCQ (More than one correct option)

• Very short answer type questions

• Short answer type questions

• Long answer type questions

Ques. A particle is dropped from a height $H$. The de-Broglie wavelength of the particle as a function of height is proportional to

Thinking process :

The de-broglie wavelength ${\lambda}$ is given by ${\lambda}=\frac{h}{m v}$

Ans. (d)

Velocity of a body falling from a height $H$ is given by

$v=\sqrt{2 g H}$

We know that de-broglie wavelength

Sol. to be continued …

$\lambda=\frac{h}{m v}=\frac{h}{m \sqrt{2 g H}}$

$\Rightarrow\frac{h}{m \sqrt{2 g} \sqrt{H}}$

Here, $\frac{h}{m \sqrt{2 g}}$ is a constant $\phi$ say ‘$K$’.

So,

$\lambda=K \frac{1}{\sqrt{H}} \Rightarrow \lambda \propto \frac{1}{\sqrt{H}}$

$\lambda \propto H^{-1 / 2}$

Ques. The wavelength of a photon needed to remove a proton from a nucleus which is bound to the nucleus with 1 MeV energy is nearly

Thinking process :
Energy of a photon is $E=\frac{h c}{\lambda}$, where $\lambda$ is the minimum wavelength of the photon required to eject the proton from nucleus.

Ans. (b)

Given in the question,

Energy of a photon, $E=1 MeV$

$\Rightarrow10^{6} eV$

Now,

Sol. to be continued …

$h c=1240 eVnm$

Now,

$E=\frac{h c}{\lambda}$

$\Rightarrow$ $l=\frac{h c}{E}=\frac{1240 eVnm}{10^{6} eV}$

$=1.24 \times 10^{-3} nm$

Ques. Consider a beam of electrons (each electron with energy E$_{0}$) incident on a metal surface kept in an evacuated chamber. Then,

Ans. (d)

When a beam of electrons of energy $E_0$ is incident on a metal surface kept in an evacuated chamber electrons can be emitted with maximum energy $E_0$ (due to elastic collision) and with any energy less than $E_0$, when part of incident energy of electron is used in liberating the electrons from the surface of metal.

Ques. Consider figure given below. Suppose the voltage applied to $A$ is increased. The diffracted beam will have the maximum at a value of $\theta$ that

Thinking process :

The figure given here shows the Davisson-Germer experiment which was held to verify the wave nature of electrons.

Sol. to be continued …

Ans. (c)

In Davisson-Germer experiment, the de-Broglie wavelength associated with electron is

$\lambda=\frac{12.27}{\sqrt{V}} \AA$

where $V$ is the applied voltage.

If there is a maxima of the diffracted electrons at an angle $\theta$, then

$2 d \sin \theta=\lambda$

Sol. to be continued …

From(i.), we note that if $V$ is inversely proportional to the wavelength $\lambda$. i.e., $V$ will increase with the decrease in the $\lambda$.

From Eq. (ii), we note that wavelength $\lambda$ is directly proportional to $\sin \theta$ and hence $\theta$.

So, with the decrease in $\lambda, \theta$ will also decrease.

Thus, when the voltage applied to $A$ is increased.

The diffracted beam will have the maximum at a value of $\theta$ that will be less than the earlier value.

Ques. A proton, a neutron, an electron and an $\alpha$-particle have same energy. Then, their de-Broglie wavelengths compare as

Thinking process :

The energy of any particle can be given by $K=\frac{1}{2} m v^{2}$
$\Rightarrow \quad m v=\sqrt{2 m k}$

Also, de-Broglie wavelength is given by

$\lambda=\frac{h}{m v}$

Now, relation between energy and wavelength of any particle is given by putting the value of Eq. (i) in Eq. (ii)
$\lambda=\frac{h}{\sqrt{2 m k}}$

Sol. to be continued …

Ans. (b)

We know that the relation between $\lambda$ and $K$ is given by

$\lambda=\frac{h}{\sqrt{2 m k}}$

Here, for the given value of energy $K, \frac{h}{\sqrt{2 k}}$ is a constant.

Thus, $\quad \lambda \propto \frac{1}{\sqrt{m}}$

Sol. to be continued …

$\therefore \lambda_p: \lambda_n: \lambda_e: \lambda _{\alpha}$

$\Rightarrow \frac{1}{\sqrt{m_p}}: \frac{1}{\sqrt{m_n}}: \frac{1}{\sqrt{m_e}}: \frac{1}{\sqrt{m _{\alpha}}}$

Since, $m_p=m_n$, hence ${\lambda} _{p}=\lambda_n$

As, $m _{\alpha}>m_p$, therefore $\lambda _{\alpha}<\lambda_p$

As, $ m_e\lambda_n$

Hence, $\lambda _{\alpha}<\lambda_p=\lambda_n<\lambda_e$

Ques. An electron is moving with an initial velocity ${v}=v_0 \hat{{i}}$ and is in a magnetic field ${B}=B_0 \hat{{j}}$. Then, it’s de-Broglie wavelength

Ans. (a)

Given,

$\mathbf{v}=v_0 \hat{{i}} \Rightarrow {B}=B_0 \hat{{j}}$

Force on moving electron due to magnetic field is,

$F=-e(v \times B)$

$=-e[v_0 {i} \times B_0 {j}]$

Sol. to be continued …

$\Rightarrow -e v_0 B_0 {k}$

As this force is perpendicular to ${v}$ and ${B}$, so the magnitude of ${v}$ will not change,

i.e., momentum $(=m v)$ will remain constant in magnitude. Hence,

de-Broglie wavelength $\lambda=\frac{h}{m v}$ remains constant.

Ques. An electron (mass $m$) with an initial velocity $\mathbf{v}=v_0 \hat{\mathbf{i}}(v_0>0)$ is in an electric field $\mathbf{E}=-E_0 \hat{\mathbf{i}}(E_0=$ constant $>0).$ It’s de-Broglie wavelength at time $t$ is given by

Thinking process :

de - Broglie wavelength is given by ${\lambda}=\frac{h}{m v}$

Ans. (a)

Initial de-Broglie wavelength of electron,

$\lambda_0=\frac{h}{m v_0}$

Force on electron in electric field,

Sol. to be continued …

$\mathbf{F}=-e \mathbf{E}=-e[-E_0 \hat{\mathbf{i}}]=e E_0 \hat{\mathbf{i}}$

Acceleration of electron

$a=\frac{\mathbf{F}}{m}=\frac{e E_0 \hat{\mathbf{i}}}{m}$

Velocity of electron after time $t$,

$v =v_0 \hat{\mathbf{i}}+\frac{e E_0 \hat{\mathbf{i}}}{m} t=v_0+\frac{e E_0}{m} t \hat{\mathbf{i}}$

$=v_0 1+\frac{e E_0}{m v_0} t \hat{\mathbf{i}}$

Sol. to be continued …

de - Broglie wavelength associated with electron at time $t$ is

$\lambda=\frac{h}{m v}$

$=\frac{h}{m v_0 1+\frac{e E_0}{m v_0} t}=\frac{\lambda_0}{1+\frac{e E_0}{m v_0} t}$

$\because \lambda_0=\frac{h}{m v_0}$

Ques. An electron (mass $m$ ) with an initial velocity $\mathbf{v}=v_0 \hat{\mathbf{i}}$ is in an electric field $\mathbf{E}=E_0 \hat{\mathbf{j}}$. If $\lambda_0=h / m v_0$, it’s de-Broglie wavelength at time $t$ is given by

Ans. (c)

Initial de - Broglie wavelength of electron,

$\lambda_0=\frac{h}{m v_0}$

Force on electron in electric field,

$F=-e E=-e E_0 \hat{j}$

Acceleration of electron,

Sol. to be continued …

$a=\frac{F}{m}=\frac{e E_0 \hat{j}}{m}$

It is acting along negative $y$-axis.

The initial velocity of electron along $x$-axis, $x_0=v_0 \hat{\mathbf{i}}$. Initial velocity of electron along $y$-axis,

$y_0=0 .$

Velocity of electron after time $t$ along $x$-axis, $x =_0 \hat{\mathbf{i}}$

Velocity of electron after timet along $y$-axis,

Sol. to be continued …

$y=0+-\frac{e E_0}{m} \hat{{j}} t=-\frac{e E_0}{m} t \hat{{j}}$
Magnitude of velocity of electron after time $t$ is

$v =\sqrt{v_x^{2}+v_y^{2}}=\sqrt{v_0^{2}+\frac{-e E_0}{m} t^{2}}$

$\Rightarrow =v_0 \sqrt{1+\frac{e^{2} E_0^{2} t^{2}}{m^{2} v_0^{2}}}$

de-Broglie wavelength, $\lambda^{\prime} =\frac{h}{m v}$

$=\frac{h}{m v_0 \sqrt{1+e^{2} E_0^{2} t^{2} /(m^{2} v_0^{2})}}$

$=\frac{\lambda_0}{\sqrt{1+e^{2} E_0^{2} t^{2} / m^{2} v_0^{2}}}$

Ques. Relativistic corrections become necessary when the expression for the kinetic energy $\frac{1}{2} m v^{2}$, becomes comparable with $m c^{2}$, where $m$ is the mass of the particle. At what de-Broglie wavelength, will relativistic corrections become important for an electron?

Thinking process :

The de-Broglie wavelength at which relativistic corrections become important must be greater than speed of light i.e., $3 \times 10^{8} m / s$.

Ans. (c, d)

de-Broglie wavelength

Here,

$\lambda=\frac{h}{m v} \Rightarrow v=\frac{h}{m \lambda}$

Sol. to be continued …

$h=6.6 \times 10^{-34} Js$

and for electron,

Now consider each option one by one

(a) $\lambda_1=10 nm=10 \times 10^{-9} m=10^{-8} m$

$\Rightarrow v_1=\frac{6.6 \times 10^{-34}}{(9 \times 10^{-31}) \times 10^{-8}}$

$\Rightarrow \frac{2.2}{3} \times 10^{5} \approx 10^{5}$ m/s

Sol. to be continued …

(b) $\lambda_2=10^{-1} nm=10^{-1} \times 10^{-9} m=10^{-10} m$

$\Rightarrow \quad v_2=\frac{6.6 \times 10^{-34}}{(9 \times 10^{-31}) \times 10^{-10}} \approx 10^{7} m / s$

(c) $\lambda_3=10^{-4} nm=10^{-4} \times 10^{-9} m=10^{-13} m$

$\Rightarrow \quad v_3=\frac{6.6 \times 10^{-34}}{(9 \times 10^{-31}) \times 10^{-13}} \approx 10^{10} m / s$

(d) $\lambda_4=10^{-6} nm=10^{-6} \times 10^{-9} m=10^{-15} m$

$\Rightarrow \quad v_4=\frac{6.6 \times 10^{-34}}{9 \times 10^{-31} \times 10^{-15}} \approx 10^{12} m / s$

Thus, options (c) and (d) are correct as $v_3$ and $v_4$ is greater than $3 \times 10^{8} m / s$.

Ques. Two particles $A_1$ and $A_2$ of masses $m_1, m_2(m_1>m_2)$ have the same de-Broglie wavelength. Then,

Ans. (a, c)

de-Broglie wavelength $\lambda=\frac{h}{m v}$

where, $m v=p$ (moment)

$\Rightarrow$ $\lambda=\frac{h}{p} \Rightarrow p=\frac{h}{\lambda}$

Here, $h$ is a constant.

So,

Sol. to be continued …

$p \propto \frac{1}{\lambda} \Rightarrow \frac{p_1}{p_2}=\frac{\lambda_2}{\lambda_1}$

$(\lambda_1=\lambda_2)=\lambda$

$\frac{p_1}{p_2}=\frac{\lambda}{\lambda}=1 \Rightarrow p_1=p_2$

But

Then,

Thus, their momenta is same.

Sol. to be continued …

Also,

$E =\frac{1}{2} m v^{2}=\frac{1}{2} \frac{m v^{2} \times m}{m}$

$=\frac{1}{2} \frac{m^{2} v^{2}}{m}=\frac{1}{2} \frac{p^{2}}{m}$

Here, $p$ is constant

$E \propto \frac{1}{m}$

$\frac{E_1}{E_2}=\frac{m_2}{m_1}<1 \Rightarrow E_1

Ques. The de-Broglie wavelength of a photon is twice, the de-Broglie wavelength of an electron. The speed of the electron is $v_e=\frac{c}{100}$. Then,

Ans. (b, c)

Suppose, Mass of electron $=m_e,$ Mass of photon $=m_p$,

Velocity of electron $=v_e$ and Velocity of photon $=v_p$

Thus, for electron, de-Broglie wavelength

$\lambda_e =\frac{h}{m_e v_e}$

$=\frac{h}{m_e(c / 100)}=\frac{100 h}{m_e c} \text{ (given) }$

Sol. to be continued …

Kinetic energy,

$E_e=\frac{1}{2} m_e v_e^{2}$

$\Rightarrow \quad m_e v_e=\sqrt{2 E_e m_e}$

$\text{ so, } \quad \lambda_e=\frac{h}{m_e v_e}=\frac{h}{\sqrt{2 m_e E_e}}$

$\Rightarrow \quad E_e=\frac{h^{2}}{2 \lambda_e^{2} m_e}$

For photon of wavelength $\lambda_p$, energy

Sol. to be continued …

$E_p=\frac{h c}{\lambda_p}=\frac{h c}{2 \lambda_e}$

$\therefore \quad \frac{E_p}{E_e}=\frac{h c}{2 \lambda_e} \times \frac{2 \lambda_e^{2} m_e}{h^{2}}$

$=\frac{\lambda_e m_e c}{h}=\frac{100 h}{m_e c} \times \frac{m_e c}{h}=100$

So,

For electron,

So,

Sol. to be continued …

$\frac{E_e}{E_p}=\frac{1}{100}=10^{-2}$

$p_e=m_e v_e=m_e \times c / 100$

$\frac{p_e}{m_e c}=\frac{1}{100}=10^{-2}$

Ques. Photons absorbed in matter are converted to heat. A source emitting $n$ photon/sec of frequency $v$ is used to convert 1 kg of ice at $0^{\circ} C$ to water at $0^{\circ} C$. Then, the time $T$ taken for the conversion

Ans. (a, b, c)

Energy spent to convert ice into water

$=\text{ mass } \times \text{ latent heat }$

$=m L=(1000 g) \times(80 cal / g)$

$=80000 cal$

Energy of photons used $=n T \times E=n T \times h \nu$

Sol. to be continued …

So,

$n T h \nu=m L \Rightarrow T=\frac{m L}{n h v}$

$\therefore \quad T \propto \frac{1}{n}$, when $v$ is constant.

$T \propto \frac{1}{v}$, when $n$ is fixed.

$\Rightarrow \quad T \propto \frac{1}{n v}$

Thus, $T$ is constant, if $n v$ is constant.

Ques. A particle moves in a closed orbit around the origin, due to a force which is directed towards the origin. The de-Broglie wavelength of the particle varies cyclically between two values $\lambda_1, \lambda_2$ with $\lambda_1>\lambda_2$. Which of the following statement are true?

Ans. (b, d)

The de-Broglie wavelength of the particle can be varying cyclically between two values $\lambda_1$ and $\lambda_2$, if particle is moving in an elliptical orbit with origin as its one focus.

Consider the figure given below

Let $v_1, v_2$ be the speed of particle at $A$ and $B$ respectively and origin is at focus $O$. If $\lambda_1, \lambda_2$ are the de-Broglie wavelengths associated with particle while moving at $A$ and $B$ respectively. Then,
and

$\lambda_1=\frac{h}{m v_1}$

$\lambda_2=\frac{h}{m v_2}$

$\therefore \frac{\lambda_1}{\lambda_2}=\frac{v_2}{v_1}$

$\text{ since } \lambda_1>\lambda_2$

Sol. to be continued …

$\therefore v_2>v_1$

By law of conservation of angular momentum, the particle moves faster when it is closer to focus.

From figure, we note that origin $O$ is closed to $P$ than $A$.

Ques. A proton and an $\alpha$-particle are accelerated, using the same potential difference. How are the de-Broglie wavelengths $\lambda_p$ and $\lambda _a$ related to each other?

Thinking process :

Here, since both proton and ${\alpha}$-particle use the same potential difference, thus they are taken as constant.

Ans.

As,

$\lambda =\frac{h}{\sqrt{2 m q v}}$

$\lambda \propto \frac{1}{\sqrt{m q}}$

Sol. to be continued …

$\frac{\lambda_p}{\lambda _{\alpha}} =\frac{\sqrt{m _{\alpha} q _{\alpha}}}{\sqrt{m_p q_p}}=\frac{\sqrt{4 m_p \times 2 e}}{\sqrt{m_p \times e}}=\sqrt{8}$

$\therefore \quad \lambda_p=\sqrt{8} \lambda _{\alpha}$

i.e., wavelength of proton is $\sqrt{8}$ times wavelength of $\alpha$-particle.

Ques. (i) In the explanation of photoeletric effect, we assume one photon of frequency $v$ collides with an electron and transfers its energy. This leads to the equation for the maximum energy $E _{\max }$ of the emitted electron as

$E _{\max }=h \nu-\phi_0$

where $\phi_0$ is the work function of the metal. If an electron absorbs 2 photons (each of frequency $\nu$ ), what will be the maximum energy for the emitted electron?

(ii) Why is this fact (two photon absorption) not taken into consideration in our discussion of the stopping potential?

Ans.

(i) Here it is given that, an electron absorbs 2 photons each of frequency $v$ then $v^{\prime}=2 v$ where, $v^{\prime}$ is the frequency of emitted electron.

Given,

$E _{\max }=h \nu-\phi_0$

Now, maximum energy for emitted electrons is

$E _{\max }^{\prime}=h(2 v)-\phi_0=2 h \nu-\phi_0$

(ii) The probability of absorbing 2 photons by the same electron is very low. Hence, such emission will be negligible.

Ques. There are materials which absorb photons of shorter wavelength and emit photons of longer wavelength. Can there be stable substances which absorb photons of larger wavelength and emit light of shorter wavelength.

Ans.

According to first statement, when the materials which absorb photons of shorter wavelength has the energy of the incident photon on the material is high and the energy of emitted photon is low when it has a longer wavelength.

But in second statement, the energy of the incident photon is low for the substances which has to absorb photons of larger wavelength and energy of emitted photon is high to emit light of shorter wavelength.

This means in this statement material has to supply the energy for the emission of photons.

But this is not possible for a stable substances.

Ques. Do all the electrons that absorb a photon come out as photoelectrons?

Ans.

In photoelectric effect, we can observe that most electrons get scattered into the metal by absorbing a photon.

Thus, all the electrons that absorb a photon doesn’t come out as photoelectron. Only a few come out of metal whose energy becomes greater than the work function of metal.

Ques. There are two sources of light, each emitting with a power of 100 W. One emits $X$-rays of wavelength 1 nm and the other visible light at 500 nm. Find the ratio of number of photons of $X$-rays to the photons of visible light of the given wavelength?

Ans.

Suppose wavelength of $X$-rays is $\lambda_1$ and the wavelength of visible light is $\lambda_2$.

Given, $\quad P=100 W$

and

$\lambda_1=1 nm$

$\lambda_2=500 nm$

Sol. to be continued …

Also, $n_1$ and $n_2$ represents number of photons of X-rays and visible light emitted from the two sources per sec.

So,

$\frac{E}{t}=P=n_1 \frac{h c}{\lambda_1}=n_2 \frac{h c}{\lambda_2}$

$\Rightarrow$

$\frac{n_1}{\lambda_1}=\frac{n_2}{\lambda_2}$

$\Rightarrow$

$\frac{n_1}{n_2}=\frac{\lambda_1}{\lambda_2}=\frac{1}{500}$

Ques. Consider figure for photoemission. How would you reconcile with momentum-conservation? Note light (photons) have momentum in a different direction than the emitted electrons.

Ans.

During photoelectric emission, the momentum of incident photon is transferred to the metal. At microscopic level, atoms of a metal absorb the photon and its momentum is transferred mainly to the nucleus and electrons.

The excited electron is emitted. Therefore, the conservation of momentum is to be considered as the momentum of incident photon transferred to the nucleus and electrons.

Ques. Consider a metal exposed to light of wavelength 600 nm. The maximum energy of the electron doubles when light of wavelength 400 nm is used. Find the work function in eV.

Thinking process :

Maximum energy $=h \nu-\phi$

Ans.

Given,

For the first condition,

Wavelength of light $\lambda=600 nm$

Sol. to be continued …

and for the second condition,

Wavelength of light $\lambda^{\prime}=400 nm$

Also, maximum kinetic energy for the second condition is equal to the twice of the kinetic energy in first condition.

$\text{ i.e., } K _{\max }^{\prime} =2 K _{\max }$

$\text{ Here, } K _{\max }^{\prime} =\frac{h c}{\lambda}-\phi$

$\Rightarrow 2 K _{\max } =\frac{h c}{\lambda^{\prime}}-\phi_0$

Sol. to be continued …

$\Rightarrow 2 \frac{1230}{600}-\phi =\frac{1230}{400}-\phi$

$\Rightarrow \phi =\frac{1230}{1200}=1.02 eV$

Ques. Assuming an electron is confined to a $1 nm$ wide region, find the uncertainty in momentum using Heisenberg uncertainty principle $(\Delta x \times \Delta p \approx h)$. You can assume the uncertainty in position $\Delta x$ as $1 nm$. Assuming $p \approx \Delta p$, find the energy of the electron in electronvolts.

Ans.

Here, $\Delta x=1 nm=10^{-9} m, \Delta p=$ ?

As $\Delta x \Delta p \approx h$

$\therefore \Delta p =\frac{h}{\Delta x}=\frac{h}{2 \pi \Delta x}$

$\Rightarrow \frac{6.62 \times 10^{-34} Js}{2 \times(22 / 7)(10^{-9}) m}$

$=1.05 \times 10^{-25} kg m / s$

Sol. to be continued …

$\text{ Energy, } E =\frac{p^{2}}{2 m}=\frac{(\Delta p)^{2}}{2 m}$

$=\frac{(1.05 \times 10^{-25})^{2}}{2 \times 9.1 \times 10^{-31}} J$

$\Rightarrow =\frac{(1.05 \times 10^{-25})^{2}}{2 \times 9.1 \times 10^{-31} \times 1.6 \times 10^{-19}} eV$

$=3.8 \times 10^{-2} eV$

Ques. Two monochromatic beams $A$ and $B$ of equal intensity $I$, hit a screen. The number of photons hitting the screen by beam $A$ is twice that by beam $B$. Then, what inference can you make about their frequencies?

Ans.

Suppose $n_A$ is the number of photons falling per second of beam $A$ and $n_B$ is the number of photons falling per second of beam $B$.

Thus,

$n_A=2 n_B$

$A=h v_A$

$B=h v_B$

Sol. to be continued …

Energy of falling photon of beam Energy of falling photon of beam

Now, according to question,

intensity of $A=$ intensity of $B$

$\therefore \quad n_A h \nu_A=n_B h v_E$

$\Rightarrow \quad \frac{v_A}{v_B}=\frac{n_B}{n_A}=\frac{n_B}{2 n_B}=\frac{1}{2}$

$\Rightarrow \quad v_B=2 v_A$

Thus, from this relation we can infer that frequency of beam $B$ is twice of beam $A$.

Ques. Two particles $A$ and $B$ of de-Broglie wavelengths $\lambda_1$ and $\lambda_2$ combine to form a particle $C$. The process conserves momentum. Find the de-Broglie wavelength of the particle $C$. (The motion is one-dimensional)

Ans.
Given from conservation of momentum,

$|C|=|A|+|B| \frac{h}{\lambda_C}=\frac{h}{\lambda_A}+\frac{h}{\lambda_B} \quad \newline \because \lambda=\frac{h}{m v}=\frac{h}{p} \Rightarrow p=\frac{h}{\lambda} \newline$

$\Rightarrow \frac{h}{\lambda_C}=\frac{h \lambda_B+h \lambda_A}{\lambda_A \lambda_B} \newline$

$\Rightarrow \frac{\lambda_C}{h}=\frac{\lambda_A \lambda_B}{h \lambda_A+h \lambda_B} \Rightarrow \lambda_C=\frac{\lambda_A \lambda_B}{\lambda_A+\lambda_B}$

$\text{Case I}$ Suppose both $p_A$ and $p_B$ are positive, then

Sol. to be continued …

$\lambda_C=\frac{\lambda_A \lambda_B}{\lambda_A+\lambda_B}$

$\text{Case II}$ When both $p_A$ and $p_B$ are negative, then

$\lambda_C=\frac{\lambda_A \lambda_B}{\lambda_A+\lambda_B}$

$\text{Case III}$ When $p_A>0, p_B<0$ i.e., $p_A$ is positive and $p_B$ is negative,

$\frac{h}{\lambda_C}=\frac{h}{\lambda_A}-\frac{h}{\lambda_B}=\frac{(\lambda_B-\lambda_A) h}{\lambda_A \lambda_B}$

$\Rightarrow \quad \lambda_C=\frac{\lambda_A \lambda_B}{\lambda_B-\lambda_A}$

Sol. to be continued …

$\text{Case IV}$ $\quad$ $p_A<0, p_B>0$, i.e., $p_A$ is negative and $p_B$ is positive,

$\therefore \frac{h}{\lambda_C} =\frac{-h}{\lambda_A}+\frac{h}{\lambda_B} \newline$

$\Rightarrow =\frac{(\lambda_A-\lambda_B) h}{\lambda_A \lambda_B} \Rightarrow \lambda_C=\frac{\lambda_A \lambda_B}{\lambda_A-\lambda_B}$

Ques. A neutron beam of energy $E$ scatters from atoms on a surface with a spacing $d=0.1 nm$. The first maximum of intensity in the reflected beam occurs at $\theta=30^{\circ}$. What is the kinetic energy $E$ of the beam in eV?

Ans.

Given, $d=0.1 nm$,

Now, according to Bragg’s law

$\Rightarrow 2 d \sin \theta =n \lambda \Rightarrow 2 \times 0.1 \times \sin 30=1 \lambda \newline$

$\text{ Now, } \lambda =0.1 nm \Rightarrow=10^{-10} m \newline$

$\Rightarrow \lambda =\frac{h}{m v}=\frac{h}{p} \newline$

Sol. to be continued …

$\Rightarrow p=\frac{h}{\lambda} =\frac{6.62 \times 10^{-34}}{10^{-10}} \newline$

$\text{ Now, } =6.62 \times 10^{-24} kg-m / s \newline$

$KE =\frac{1}{2} m v^{2}=\frac{1}{2} \frac{m^{2} v^{2}}{m}=\frac{1}{2} \frac{p^{2}}{m} \newline$

$=\frac{1}{2} \times \frac{(6.62 \times 10^{-24})^{2}}{1.67 \times 10^{-27}} J \newline$

$=0.21 eV$

Ques. Consider a thin target ($10^{-2}$ m square, $10^{-3}$ m thickness) of sodium, which produces a photocurrent of 100 $\mu A$ when a light of intensity 100 $W/m^{2}$ $(\lambda=660 nm)$ falls on it. Find the probability that a photoelectron is produced when a photon strikes a sodium atom.

[Take density of Na = 0.97 kg/m$^{3}$].

Thinking process :

Absorption of two photons by an atom depends on the probability of photoemission by a single photon on a single atom.

Ans.

Given, $\Rightarrow$

Intensity,

$A =10^{-2} m^{2}=10^{-2} \times 10^{-2} m^{2}$

Sol. to be continued …

$=10^{-4} m^{2}$

$d =10^{-3} m$

$i =100 \times 10^{-6} A=10^{-4} A$

$I =100 W / m^{2}$

$\lambda =660 nm=660 \times 10^{-9} m$

$\text{ Pa } =0.97 kg / m^{3}$

Sol. to be continued …

$\text{ aber } =6 \times 10^{26} kg \text{ atom }$

$\text{ rget } =A \times d$

$=10^{-4} \times 10^{-3}$

$=10^{-7} m^{3}$

$\text{ Avogadro’s number } =6 \times 10^{26} kg \text{ atom }$

Volume of sodium target = $A \times d$

Sol. to be continued …

We know that $6 \times 10^{26}$ atoms of Na weights $=23 kg$

So, volume of $6 \times 10^{6} Na$ atoms $=\frac{23}{0.97} m^{3}$

Volume occupied by one $Na$ atom $=\frac{23}{0.97 \times(6 \times 10^{26})}=3.95 \times 10^{-26} m^{3}$