Thinking Process

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

Answer

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

$$ v=\sqrt{2 g H} $$

We know that de-broglie wavelength

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

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

So,

$$ \begin{aligned} & \lambda=K \frac{1}{\sqrt{H}} \Rightarrow \lambda \propto \frac{1}{\sqrt{H}} \\ & \lambda \propto H^{-1 / 2} \end{aligned} $$

Answer

(b) Given in the question,

Energy of a photon, $E=1 MeV \Rightarrow=10^{6} eV$

Now,

$h c=1240 eVnm$

Now,

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

$\Rightarrow$

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

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

Answer

(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.

Thinking Process

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

Answer

(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

$$ 2 d \sin \theta=\lambda $$

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.

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)

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

Answer

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

$\therefore \quad \lambda_{p}: \lambda_{n}: \lambda_{e}: \lambda_{\alpha}$

$\Rightarrow \quad=\frac{1}{\sqrt{m_{p}}}: \frac{1}{\sqrt{m_{n}}}: \frac{1}{\sqrt{m_{e}}}: \frac{1}{\sqrt{m_{\alpha}}}$

Since, $\quad m_{p}=m_{n}$, hence $\boldsymbol{\lambda}{p}=\lambda{n}$

As, $\quad m_{\alpha}>m_{p}$, therefore $\lambda_{\alpha}<\lambda_{p}$

As, $\quad m_{e}<m_{n}$, therefore $\lambda_{e}>\lambda_{n}$

Hence, $\quad \lambda_{\alpha}<\lambda_{p}=\lambda_{n}<\lambda_{e}$

Answer

(a) Given,

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

Force on moving electron due to magnetic field is, $F=-e(v \times B)$

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

As this force is perpendicular to $\mathbf{v}$ and $\mathbf{B}$, so the magnitude of $\mathbf{v}$ will not change, i.e., momentum $(=m v)$ will remain constant in magnitude. Hence,

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

Thinking Process

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

Answer

(a) Initial de-Broglie wavelength of electron,

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

Force on electron in electric field,

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

Acceleration of electron

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

Velocity of electron after time $t$,

$$ \begin{aligned} v & =v_0 \hat{\mathbf{i}}+\frac{e E_0 \hat{\mathbf{i}}}{m} t=v_0+\frac{e E_0}{m} t \hat{\mathbf{i}} \\ & =v_0 1+\frac{e E_0}{m v_0} t \hat{\mathbf{i}} \end{aligned} $$

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

$$ \begin{aligned} & \lambda=\frac{h}{m v} \\ & =\frac{h}{m v_0 1+\frac{e E_0}{m v_0} t}=\frac{\lambda_0}{1+\frac{e E_0}{m v_0} t} \\ & \because \lambda_0=\frac{h}{m v_0} \end{aligned} $$

Answer

(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,

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

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,

$$ y=0+-\frac{e E_0}{m} \hat{\mathbf{j}} t=-\frac{e E_0}{m} t \hat{\mathbf{j}} $$

Magnitude of velocity of electron after time $t$ is

$$ v =\sqrt{v_x^{2}+v_y^{2}}=\sqrt{v_0^{2}+\frac{-e E_0}{m} t^{2}} \newline $$ $$\Rightarrow =v_0 \sqrt{1+\frac{e^{2} E_0^{2} t^{2}}{m^{2} v_0^{2}}} \newline $$ $$\text { de-Broglie wavelength, } \lambda^{\prime} =\frac{h}{m v} \newline $$ $$ =\frac{h}{m v_0 \sqrt{1+e^{2} E_0^{2} t^{2} /(m^{2} v_0^{2})}} \newline $$ $$ =\frac{\lambda_0}{\sqrt{1+e^{2} E_0^{2} t^{2} / m^{2} v_0^{2}}} $$

Thinking Process

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

Answer

$(c, d)$

de-Broglie wavelength

Here,

$$ \begin{aligned} & \lambda=\frac{h}{m v} \Rightarrow v=\frac{h}{m \lambda} \\ & h=6.6 \times 10^{-34} Js \end{aligned} $$

and for electron,

Now consider each option one by one

(a)

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

$$ \begin{matrix} \Rightarrow & v_1=\frac{6.6 \times 10^{-34}}{(9 \times 10^{-31}) \times 10^{-8}} \\ \Rightarrow & =\frac{2.2}{3} \times 10^{5} \approx 10^{5} m / s \end{matrix} $$

(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 $3 \times 10^{8} m / s$.

Answer

(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,

$$ \begin{gathered} p \propto \frac{1}{\lambda} \Rightarrow \frac{p_1}{p_2}=\frac{\lambda_2}{\lambda_1} \\ (\lambda_1=\lambda_2)=\lambda \\ \frac{p_1}{p_2}=\frac{\lambda}{\lambda}=1 \Rightarrow p_1=p_2 \end{gathered} $$

But

Then,

Thus, their momenta is same.

Also,

$$ \begin{aligned} E & =\frac{1}{2} m v^{2}=\frac{1}{2} \frac{m v^{2} \times m}{m} \\ & =\frac{1}{2} \frac{m^{2} v^{2}}{m}=\frac{1}{2} \frac{p^{2}}{m} \end{aligned} $$

Here, $p$ is constant

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

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

Answer

( $b, c$ )

Suppose, Mass of electron $=m_{e}, \quad$ Mass of photon $=m_{p}$,

Velocity of electron $=v_{e}$ and $\quad$ Velocity of photon $=v_{p}$

Thus, for electron, de-Broglie wavelength

$$ \begin{aligned} \lambda_{e} & =\frac{h}{m_{e} v_{e}} \\ & =\frac{h}{m_{e}(c / 100)}=\frac{100 h}{m_{e} c} \text { (given) } \end{aligned} $$

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

$$ E_{p}=\frac{h c}{\lambda_{p}}=\frac{h c}{2 \lambda_{e}} $$

$$ \begin{aligned} & \therefore \quad \frac{E_{p}}{E_{e}}=\frac{h c}{2 \lambda_{e}} \times \frac{2 \lambda_e^{2} m_{e}}{h^{2}} \\ & =\frac{\lambda_{e} m_{e} c}{h}=\frac{100 h}{m_{e} c} \times \frac{m_{e} c}{h}=100 \\ & \text { So, } \\ & \text { For electron, } \\ & \text { So, } \\ & \frac{E_{e}}{E_{p}}=\frac{1}{100}=10^{-2} \\ & p_{e}=m_{e} v_{e}=m_{e} \times c / 100 \\ & \frac{p_{e}}{m_{e} c}=\frac{1}{100}=10^{-2} \end{aligned} $$

Answer

$(a, b, c)$

Energy spent to convert ice into water

$$ \begin{aligned} & =\text { mass } \times \text { latent heat } \\ & =m L=(1000 g) \times(80 cal / g) \\ & =80000 cal \end{aligned} $$

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

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.

Answer

($b, d)$

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

$$ \begin{aligned} & \lambda_1=\frac{h}{m v_1} \\ & \lambda_2=\frac{h}{m v_2} \end{aligned} $$

$$ \begin{matrix} \therefore & \frac{\lambda_1}{\lambda_2}=\frac{v_2}{v_1} \\ \text { since } & \lambda_1>\lambda_2 \\ \therefore & v_2>v_1 \end{matrix} $$

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 $A$.

Thinking Process

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

Answer

As,

$$ \begin{aligned} \lambda & =\frac{h}{\sqrt{2 m q v}} \\ \lambda & \propto \frac{1}{\sqrt{m q}} \\ \frac{\lambda_{p}}{\lambda_{\alpha}} & =\frac{\sqrt{m_{\alpha} q_{\alpha}}}{\sqrt{m_{p} q_{p}}}=\frac{\sqrt{4 m_{p} \times 2 e}}{\sqrt{m_{p} \times e}}=\sqrt{8} \end{aligned} $$

$$ \therefore \quad \lambda_{p}=\sqrt{8} \lambda_{\alpha} $$

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

Answer

(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, $\quad E_{\max }=h \nu-\phi_0$

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.

Answer

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.

Answer

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.

Answer

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

Given, $\quad P=100 W$

and

$\lambda_1=1 nm$

$\lambda_2=500 nm$

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

So,

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

$\Rightarrow$

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

$\Rightarrow$

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

Answer

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.

Answer

Given,

For the first condition,

Wavelength of light $\lambda=600 nm$

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.

$$ \begin{matrix} \text { i.e., } & K_{\max }^{\prime} & =2 K_{\max } \\ \text { Here, } & K_{\max }^{\prime} & =\frac{h c}{\lambda}-\phi \\ \Rightarrow & 2 K_{\max } & =\frac{h c}{\lambda^{\prime}}-\phi_0 \\ \Rightarrow & 2 \frac{1230}{600}-\phi & =\frac{1230}{400}-\phi \\ \Rightarrow & \phi & =\frac{1230}{1200}=1.02 eV \end{matrix} $$

Answer

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

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

$$ \begin{matrix} \therefore & & \Delta p & =\frac{h}{\Delta x}=\frac{h}{2 \pi \Delta x} \\ \Rightarrow & & =\frac{6.62 \times 10^{-34} Js}{2 \times(22 / 7)(10^{-9}) m} \\ & & =1.05 \times 10^{-25} kg m / s \\ \text { Energy, } & E & =\frac{p^{2}}{2 m}=\frac{(\Delta p)^{2}}{2 m} \\ & & =\frac{(1.05 \times 10^{-25})^{2}}{2 \times 9.1 \times 10^{-31}} J \\ \Rightarrow & & =\frac{(1.05 \times 10^{-25})^{2}}{2 \times 9.1 \times 10^{-31} \times 1.6 \times 10^{-19}} eV \\ & & =3.8 \times 10^{-2} eV . \end{matrix} $$

Answer

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,

$$ \begin{gathered} n_{A}=2 n_{B} \\ A=h v_{A} \\ B=h v_{B} \end{gathered} $$

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

Now, according to question,

intensity of $A=$ intensity of $B$

$$ \begin{aligned} & \therefore \quad n_{A} h \nu_{A}=n_{B} h v_{E} \\ & \Rightarrow \quad \frac{v_{A}}{v_{B}}=\frac{n_{B}}{n_{A}}=\frac{n_{B}}{2 n_{B}}=\frac{1}{2} \\ & \Rightarrow \quad v_{B}=2 v_{A} \end{aligned} $$

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

Answer

Given from conservation of momentum,

$$ \begin{matrix} |C|=|A|+|B| \frac{h}{\lambda_{C}}=\frac{h}{\lambda_{A}}+\frac{h}{\lambda_{B}} \quad \newline \because \lambda=\frac{h}{m v}=\frac{h}{p} \Rightarrow p=\frac{h}{\lambda} \newline \\ \Rightarrow \frac{h}{\lambda_{C}}=\frac{h \lambda_{B}+h \lambda_{A}}{\lambda_{A} \lambda_{B}} \newline \\ \Rightarrow \frac{\lambda_{C}}{h}=\frac{\lambda_{A} \lambda_{B}}{h \lambda_{A}+h \lambda_{B}} \Rightarrow \lambda_{C}=\frac{\lambda_{A} \lambda_{B}}{\lambda_{A}+\lambda_{B}} \end{matrix} $$

Case I Suppose both $p_{A}$ and $p_{B}$ are positive, then

$$ \lambda_{C}=\frac{\lambda_{A} \lambda_{B}}{\lambda_{A}+\lambda_{B}} $$

Case II When both $p_{A}$ and $p_{B}$ are negative, then

$$ \lambda_{C}=\frac{\lambda_{A} \lambda_{B}}{\lambda_{A}+\lambda_{B}} $$

Case III When $p_{A}>0, p_{B}<0$ i.e., $p_{A}$ is positive and $p_{B}$ is negative,

$$ \begin{aligned} & \frac{h}{\lambda_{C}}=\frac{h}{\lambda_{A}}-\frac{h}{\lambda_{B}}=\frac{(\lambda_{B}-\lambda_{A}) h}{\lambda_{A} \lambda_{B}} \\ & \Rightarrow \quad \lambda_{C}=\frac{\lambda_{A} \lambda_{B}}{\lambda_{B}-\lambda_{A}} \end{aligned} $$

Case IV $\quad$ $p_{A}<0, p_{B}>0$, i.e., $p_{A}$ is negative and $p_{B}$ is positive,

$$ \begin{matrix} \therefore \frac{h}{\lambda_{C}} =\frac{-h}{\lambda_{A}}+\frac{h}{\lambda_{B}} \newline \\ \Rightarrow =\frac{(\lambda_{A}-\lambda_{B}) h}{\lambda_{A} \lambda_{B}} \Rightarrow \lambda_{C}=\frac{\lambda_{A} \lambda_{B}}{\lambda_{A}-\lambda_{B}} \end{matrix} $$

Answer

Given, $d=0.1 nm$,

Now, according to Bragg’s law

$$ \begin{matrix} \Rightarrow 2 d \sin \theta =n \lambda \Rightarrow 2 \times 0.1 \times \sin 30=1 \lambda \newline \\ \text { Now, } \lambda =0.1 nm \Rightarrow=10^{-10} m \newline \\ \Rightarrow \lambda =\frac{h}{m v}=\frac{h}{p} \newline \\ \Rightarrow p=\frac{h}{\lambda} =\frac{6.62 \times 10^{-34}}{10^{-10}} \newline \\ \text { Now, } =6.62 \times 10^{-24} kg-m / s \newline \\ KE =\frac{1}{2} m v^{2}=\frac{1}{2} \frac{m^{2} v^{2}}{m}=\frac{1}{2} \frac{p^{2}}{m} \newline \\ =\frac{1}{2} \times \frac{(6.62 \times 10^{-24})^{2}}{1.67 \times 10^{-27}} J \newline \\ =0.21 eV \end{matrix} $$

Thinking Process

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

Answer

Given, $\Rightarrow$

Intensity, $$ \begin{aligned} A & =10^{-2} m^{2}=10^{-2} \times 10^{-2} m^{2} \\ & =10^{-4} m^{2} \\ d & =10^{-3} m \\ i & =100 \times 10^{-6} A=10^{-4} A \\ I & =100 W / m^{2} \\ \lambda & =660 nm=660 \times 10^{-9} m \\ \text { Pa } & =0.97 kg / m^{3} \\ \text { aber } & =6 \times 10^{26} kg \text { atom } \\ \text { rget } & =A \times d \\ & =10^{-4} \times 10^{-3} \\ & =10^{-7} m^{3} \end{aligned} $$

$$ \begin{aligned} \text { Avogadro’s number } & =6 \times 10^{26} kg \text { atom } \\ \text { Volume of sodium target } & =A \times d \end{aligned} $$

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 $=\frac{23}{0.97 \times(6 \times 10^{26})}=3.95 \times 10^{-26} m^{3}$

Number of $Na$ atoms in target $({ }^{n} Na)$

$$ =\frac{10^{-7}}{3.95 \times 10^{-26}}=2.53 \times 10^{18} $$

Let $n$ be the number of photons falling per second on the target.

Energy of each photon $=h c / \lambda$

Total energy falling per second on target $=\frac{n h c}{\lambda}=I A$

$$ \begin{aligned} \therefore & n & =\frac{I A \lambda}{h c} \\ \Rightarrow & & =\frac{100 \times 10^{-4} \times(660 \times 10^{-9})}{(6.62 \times 10^{-34}) \times(3 \times 10^{8})}=3.3 \times 10^{16} \end{aligned} $$

Let $P$ be the probability of emission per atom per photon.

The number of photoelectrons emitted per second

$$ \begin{aligned} N & =P \times n \times({ }^{n} Na) \\ & =P \times(3.3 \times 10^{16}) \times(2.53 \times 10^{18}) \end{aligned} $$

Now, according to question,

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

Current,

$$ \begin{aligned} i & =N e \\ 10^{-4} & =P \times(3.3 \times 10^{16}) \times(2.53 \times 10^{18}) \times(1.6 \times 10^{-19}) \\ P & =\frac{10^{-4}}{(3.3 \times 10^{16}) \times(2.53 \times 10^{18}) \times(1.6 \times 10^{-19})} \\ & =7.48 \times 10^{-21} \end{aligned} $$

$$ \begin{matrix} \therefore & 10^{-4}=P \times(3.3 \times 10^{16}) \times(2.53 \times 10^{18}) \times(1.6 \times 10^{-19}) \\ \Rightarrow & P=c \end{matrix} $$

Thus, the probability of emission by a single photon on a single atom is very much less than 1. It is due to this reason, the absorption of two photons by an atom is negligible.

Thinking Process

$$ \text { Work done by an external agency }=+\frac{1}{4 \pi \varepsilon_0} \cdot \frac{1}{4} \int_0^{\infty} \frac{q^{2}}{x^{2}} d x $$

Answer

According to question, consider the figure given below From figure, $d=0.1 nm=10^{-10} m$,

$$ F=\frac{q^{2}}{4 \times 4 \pi \varepsilon_0 d^{2}} $$

Let the electron be at distance $x$ from metallic surface. Then, force of attraction on it is

$$ F_{x}=\frac{q^{2}}{4 \times 4 \pi \varepsilon_0 x^{2}} $$

Work done by external agency in taking the electron from distance $d$ to infinity is

$$ \begin{aligned} W & =\int_1^{\infty} F_{x} d x=\int_1^{\infty} \frac{q^{2} d x}{4 \times 4 \pi \varepsilon_0} \frac{1}{x^{2}} \\ & =\frac{q^{2}}{4 \times 4 \pi \varepsilon_0} \frac{1}{d} \\ & =\frac{(1.6 \times 10^{-19})^{2} \times 9 \times 10^{9}}{4 \times 10^{-10}} J \\ & =\frac{(1.6 \times 10^{-19})^{2} \times(9 \times 10^{9})}{(4 \times 10^{-10}) \times(1.6 \times 10^{-19})} eV=3.6 eV \end{aligned} $$

Answer

(i) Given, thresholed frequency of $A$ is given by $\nu_{O A}=5 \times 10^{14} Hz$ and For $B$,

$$ \nu_{O B}=10 \times 10^{14} Hz $$

We know that

Work function,

$$ \phi=h v_0 \text { or } \phi_0 \propto v_0 $$

$\Rightarrow$

$$ \begin{aligned} \phi_0 & \propto \nu_0 \\ \frac{\phi_{D A}}{\phi_{O B}} & =\frac{5 \times 10^{14}}{10 \times 10^{14}}<1 \\ \phi_{D A} & <\phi_{O B} \end{aligned} $$

So,

$\Rightarrow$

Thus, work function of $B$ is higher than $A$.

(ii) For metal $A$, slope $=\frac{h}{e}=\frac{2}{(10-5) 10^{14}}$

or

$$ \begin{aligned} h & =\frac{2 e}{5 \times 10^{14}}=\frac{2 \times 1.6 \times 10^{-19}}{5 \times 10^{14}} \\ & =6.4 \times 10^{-34} Js \end{aligned} $$

For metal $B$, slope $=\frac{h}{e}=\frac{2.5}{(15-10) 10^{14}}$

or

$$ \begin{aligned} h & =\frac{2.5 \times e}{5 \times 10^{14}}=\frac{2.5 \times 1.6 \times 10^{-19}}{5 \times 10^{14}} \\ & =8 \times 10^{-34} Js \end{aligned} $$

Since, the value of $h$ from experiment for metals $A$ and $B$ is different. Hence, experiment is not consistent with theory.

Answer

As collision is elastic, hence laws of conservation of momentum and kinetic energy are obeyed.

According to law of conservation of momentum,

$\Rightarrow$

$$ \begin{aligned} m_{A} v+m_{B} 0 & =m_{A} v_1+m_{B} v_2 \\ m_{A}(v-v_1) & =m_{B} v_2 \end{aligned} $$

According to law of conservation of kinetic energy,

$$ \begin{matrix} \Rightarrow \frac{1}{2} m_{A} v^{2} =\frac{1}{2} m_{A} v_1^{2}+\frac{1}{2} m_{B} v_2^{2} \\ \Rightarrow m_{A}(v-v_1^{2}) =m_{B} v_2^{2} \\ \Rightarrow m_{A}(v-v_1)(v+v_1) =m_{B} v_2^{2} \end{matrix} $$

Dividing E(ii) by Eq. (i),

we get, $\quad v+v_1=v_2$ or $v=v_2-v_1$

Solving Eqs. (i) and (iii), we get

$$ \begin{aligned} v_1 & =\frac{m_{A}-m_{B}}{m_{A}+m_{B}} \quad v \text { and } v_2=\frac{2 m_{A}}{m_{A}+m_{B}} v \\ \lambda_{\text {initial }} & =\frac{h}{m_{A} v} \\ \lambda_{\text {final }} & =\frac{h}{m_{A} v_1}=\frac{h(m_{A}+m_{B})}{m_{A}(m_{A}-m_{B}) v} \\ \Delta \lambda & =\lambda_{\text {final }}-\lambda_{\text {initial }}=\frac{h}{m_{A} v} \frac{m_{A}+m_{B}}{m_{A}-m_{B}}-1 \end{aligned} $$

Answer

Given, $P=20 W, \lambda=5000 \AA=5000 \times 10^{-10} m$

$$ d=2 m, \phi_0=2 eV, r=1.5 A=1.5 \times 10^{-10} m $$

(i) Number of photon emitted by bulb per second is $n^{\prime}=\frac{p}{h c / \lambda}=\frac{p \lambda}{h c}$

$\Rightarrow$

$$ \begin{aligned} & =\frac{20 \times(5000 \times 10^{-10})}{(6.62 \times 10^{-34}) \times(3 \times 10^{8})} \\ & =5 \times 10^{19} s^{-1} \end{aligned} $$

(ii) Energy of the incident photon $=\frac{h c}{\lambda}=\frac{(6.62 \times 10^{-34})(3 \times 10^{8})}{5000 \times 10^{-10} \times 1.6 \times 10^{-19}}$

$$ =2.48 eV $$

As this energy is greater than $2 eV$ (i.e., work function of metal surface), hence photoelectric emission takes place.

(iii) Let $\Delta t$ be the time spent in getting the energy $\phi=$ (work function of metal).

Consider the figure,

$$ \Rightarrow \quad \begin{aligned} \frac{P}{4 \pi d^{2}} \times \pi r^{2} \Delta t & =\phi_0 \\ \Delta t & =\frac{4 \phi_0 d^{2}}{P r^{2}} \\ & =\frac{4 \times(2 \times 1.6 \times 10^{-19}) \times 2^{2}}{20 \times(1.5 \times 10^{-10})^{2}} \approx 28.4 s \end{aligned} $$

(iv) Number of photons received by atomic disc in time $\Delta t$ is

$$ \begin{aligned} N & =\frac{n^{\prime} \times \pi^{2}}{4 \pi d^{2}} \times \Delta t \\ & =\frac{n^{\prime} r^{2} \Delta t}{4 d^{2}} \\ & =\frac{(5 \times 10^{19}) \times(1.5 \times 10^{-10})^{2} \times 28.4}{4 \times(2)^{2}} \approx 2 \end{aligned} $$

(v) As time of emission of electrons is $11.04 s$.

Hence, the photoelectric emission is not instantaneous in this problem.

In photoelectric emission, there is an collision between incident photon and free electron of the metal surface, which lasts for very very short interval of time $(\approx 10^{-9} s)$, hence we say photoelectric emission is instantaneous