Dual Nature of Radiation

Given, wavelength of incident light, $\lambda_1=1240 \mathop A\limits^o $

Stopping potential, $V_0=8 \mathrm{~V}$

Since, energy, $E=\frac{12400}{\lambda} \mathrm{eV}$

Incident energy $E_i=\frac{12400}{1240}=10 \mathrm{~eV}$

and $E_i-e V_0=W_0$

$\therefore W_0$ (work-function of emitter)

$\begin{aligned} & =(10-8) \mathrm{~eV}=2 \mathrm{~eV} \\ & \text { and } \\ & \lambda_0=\frac{12400}{2}=6200 ~\mathop A\limits^o \\ \end{aligned}$

Given, maximum speed of photoelectron $=v_{\max }$

According to Einstein's photoelectric equation,

$\begin{aligned} & \frac{1}{2} m v_{\max }^2=h \nu-h \nu_0 \\ & \Rightarrow \quad 2_{\max } \propto \nu \\ \end{aligned}$

So, graph between $v_{\max }$ and $v$ is a parabola as shown in option $(\mathrm{c})$.

Given, electric potential, $V=100 \mathrm{~V}$

Since, $\lambda=\frac{h}{p}=\frac{h}{m c}=\frac{h}{\sqrt{2 m E}}=\frac{h}{\sqrt{2 m e V}}$

where, $\lambda$ is wavelength,

$h$ is Planck's constant i.e. $6.63 \times 10^{-34} \mathrm{Js}^{-1}$,

$m$ is mass of photon, $c$ is speed and $e$ is charge on proton

$\therefore \quad \lambda=\frac{6.63 \times 10^{-34}}{\sqrt{2 \times 1.67 \times 10^{-27} \times 1.6 \times 10^{-19} \times 200}}$

$=0.203\times10^{-11}$ m

$=2.03$ pm

It is near to 2.86 pm.

Given, wavelength of radiation $=300 \mathrm{~nm}$ $=300 \times 10^{-9} \mathrm{~m} $

Intensity $I=100 \mathrm{~W} / \mathrm{m}^2$

Number of photons falling on the surface,

$\begin{aligned} n_p & =\frac{\lambda}{h c}=\frac{300 \times 10^{-9}}{6.63 \times 10^{-34} \times 3 \times 10^8} \\ & =0.1508 \times 10^{19} \end{aligned}$

As only $2 \%$ produce photoelectrons. So, number of photoelectrons ejected,

$n_e=\frac{2}{100} \times 0.1508 \times 10^{19}=0.302 \times 10^{17}$

$\therefore$ Number of photoelectrons emitted from $2 \mathrm{~cm}^2$

$\begin{aligned} \text { area } & =2 \times 10^{-4} \times n_e \\ & =0.604 \times 10^{13}=6.04 \times 10^{12} \end{aligned}$

Minimum wavelength for X-ray is given by

${\lambda _{\min }} = {{12375} \over V}\mathop A\limits^o $

As the accelerating voltage V is decreased, $\lambda$_min increases. The intensity of emitted radiation is determined by the number of electrons bombarding the target.

$\therefore$ Accelerating voltage does not change the intensity of X-rays emitted.

The energy equation in photoelectric effect is given as

$ \begin{array}{rlrl} e V_0 & =h v-\phi_0 \\ \Rightarrow \quad & V_0 =\frac{h v}{e}-\frac{\phi_0}{e} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) \end{array} $

When the frequency is doubled,

$ \begin{aligned} V_0^{\prime} & =\frac{2 h v}{e}-\frac{\phi_0}{e} \\ \Rightarrow \quad V_0^{\prime} & =2\left(\frac{h v}{e}-\frac{\phi_0}{e}\right)+\frac{\phi_0}{e} \\ V_0^{\prime} & =2 V_0+\frac{\phi_0}{e} \quad \text { [using Eq. (i)] } \end{aligned} $

$ \Rightarrow \quad V_0^{\prime}>2 V_0 $

Thus, the stopping potential will become more than double.

Given, work-function, $\phi=2.4 \mathrm{eV}$

Energy required to ionise H -atom in ground state $=13.6 \mathrm{eV}$

According to photoelectric equation,

$ \begin{aligned} E & =\mathrm{KE}+\phi \Rightarrow \frac{h c}{\lambda}=\mathrm{KE}+\phi \\ \Rightarrow \quad \frac{1240}{\lambda} & =13.6+2.4 \\ & (\because h c=1240 \mathrm{eV}-\mathrm{nm}) \end{aligned} $

$ \begin{aligned} \Rightarrow & \frac{1240}{\lambda} =16 \\ \therefore & \lambda =\frac{1240}{16}=77.5 \mathrm{~nm} \end{aligned} $

Work-function of metal surface, $\phi_0=25 \mathrm{eV}$, Wavelength, $\lambda=310 \mathrm{~nm}$

$ \Rightarrow h c=1240 \mathrm{eV}-\mathrm{nm} $

According to Einstein's photoelectric equation,

$ \begin{aligned} e V_0 & =\frac{h c}{\lambda}-\phi_0 \\ \Rightarrow \quad e V_0 & =\frac{1240 \mathrm{eV}-\mathrm{nm}}{310 \mathrm{~nm}}-2.5 \mathrm{eV} \\ \Rightarrow \quad V_0 & =\frac{1240}{310}-25=4-25=1.5 \mathrm{~V} \end{aligned} $

In case of photoemission,

$ K_{\max }=\frac{1}{2} m v_{\max }^2=h f-\phi_0 $

For photons of energies 4 eV and 2.5 eV , we have (given, work function $=2 \mathrm{eV}$ )

$ \frac{1}{2} m v_1^2=4-2=2 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

and $\quad \frac{1}{2} m v_2^2=2.5-2=0.5\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

Dividing Eq. (i) by Eq. (ii), we get

$ \frac{v_1^2}{v_2^2}=\frac{2}{0.5}=4 \Rightarrow \frac{v_1}{v_2}=2 $

$ \text { Let } \phi_0=\text { work-function of metal. } $

Then, according to Einstein's photoelectric equation,

$ K_{\max }=\frac{1}{2} m v^2=\frac{h c}{\lambda}-\phi_0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

When wavelength is reduced by $50 \%$ i.e, $\lambda^{\prime}=\frac{\lambda}{2}$, then maximum velocity of emitted electrons is $3 v$.

$ \begin{array}{ll} \Rightarrow & \frac{1}{2} m(3 v)^2=\frac{h c}{\lambda / 2}-\phi_0 \\ \Rightarrow & 9\left(\frac{1}{2} m v^2\right)=\frac{2 h c}{\lambda}-\phi_0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{array} $

From Eqs. (i) and (ii), we have

$ \begin{aligned} & 9\left(\frac{h c}{\lambda}-\phi_0\right)=\frac{2 h c}{\lambda}-\phi_0 \\ & 7 \frac{h c}{\lambda}=8 \phi_0 \Rightarrow \phi_0=\frac{7}{8} \frac{h c}{\lambda} \end{aligned} $

Here, photon energy is given as

$ \frac{h c}{\lambda}=2.4 \mathrm{eV} $

So, work-function is

$ \phi_0=\frac{7}{8} \times 2.4=2.1 \mathrm{eV} $

Kinetic energy in photoelectric effect can be given as

KE = W $-$ W₀ [W $\to$ incident energy, W₀ $\to$ work function]

Given, (KE)₁ = 1.5 $-$ 0.5 = 1 eV

(KE)₂ = 2.5 $-$ 0.5 = 2.0 eV

$\because$ ${{{{(KE)}_1}} \over {{{(KE)}_2}}} = {1 \over {2}}$

$ \Rightarrow {{{1 \over 2}mv_1^2} \over {{1 \over 2}mv_2^2}} = {1 \over 2} \Rightarrow {{{v_1}} \over {{v_2}}} = \sqrt {{1 \over 2}} $

$\therefore$ ${v_1}:{v_2} = 1 :\sqrt 2 $

So, the ratio of maximum speeds of emitted electrons is $1 :\sqrt 2 $.