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 $2 \mathrm{~cm}^2$

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

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} $

Given that the area of the metal plate $A = 1 \times 10^{-4} m^2$, and the intensity of the radiation

$I = 16 mW/m^2 = 16 \times 10^{-3} W/m^2$.

The power $P$ falling on the metal plate is given by the product of the intensity of the radiation and the area:

$P = IA = (16 \times 10^{-3} W/m^2)(1 \times 10^{-4} m^2) = 1.6 \times 10^{-6} W$.

The energy of each incident photon is given as 10 eV. Converting this to joules (since 1 eV = $1.6 \times 10^{-19} J$):

$E_{\text{photon}} = 10 eV = 10 \times 1.6 \times 10^{-19} J = 1.6 \times 10^{-18} J$.

The number of incident photons per second $n_{\text{photon}}$ is then given by the total power divided by the energy per photon:

$n_{\text{photon}} = P / E_{\text{photon}} = 1.6 \times 10^{-6} W / 1.6 \times 10^{-18} J = 10^{12} \text{ photons/s}$.

Given that only 10% of these photons actually produce photoelectrons, the number of photoelectrons produced per second $n_{\text{electron}}$ is:

$n_{\text{electron}} = 0.1 \times n_{\text{photon}} = 0.1 \times 10^{12} = 10^{11} \text{ electrons/s}$.

The work function of the metal is given as 5 eV, which is the energy required to remove an electron from the metal. Therefore, the maximum energy of the photoelectrons $E_{\text{max}}$ is given by the energy of the incident photons minus the work function:

$E_{\text{max}} = E_{\text{photon}} - \text{Work function} = 10 eV - 5 eV = 5 eV$.

We know that $E = {{hc} \over \lambda }$

So, for atom ${E_A} = {{hc} \over {{\lambda _A}}}$

And for neutron ${E_N} = {{hc} \over {{\lambda _N}}}$

Then, ${{{E_A}} \over {{E_N}}} = {{hc} \over {{\lambda _A}}} \times {{{\lambda _N}} \over {hc}} \Rightarrow {{{\lambda _N}} \over {{\lambda _A}}}$

Here, E_A is order of eV and E_N is order of MeV.

Therefore, ${{{\lambda _N}} \over {{\lambda _A}}} = {{{E_A}} \over {{E_N}}} = {{eV} \over {MeV}} = {{1eV} \over {{{10}^6}eV}}$

${\lambda _N} = {10^{ - 6}}\,{\lambda _A}$

We know that, $\lambda = {h \over {mv}}$

From third Bohr's postulate, we have

$mvr = n{h \over {2\pi }}$

${h \over {mv}} = {{2\pi r} \over n} \Rightarrow \lambda = {{2\pi r} \over n}$

Since, $r = {a_0}{{{n^2}} \over Z}$, where a₀ is radius of Bohr's orbit having value (0.53) 1² $\mathop A\limits^o $ = 0.53 $\mathop A\limits^o $, therefore,

$\lambda = {{2\pi {a_0}{n^2}} \over {n\,.\,Z}} = {{2\pi {a_0}} \over Z}\,.\,n$

For Hydrogen Z = 1. Therefore,

${\lambda _G} = {{2\pi {a_0} \times 1} \over 1} = 2\pi {a_0}$ and ${\lambda _B} = {{2\pi {a_0} \times 3} \over 1} = 6\pi {a_0}$

Then, ${\lambda _B} = 3{\lambda _G}$