Question

In the figure shown electromagnetic radiations of wavelength $200\text{nm}$ are incident on the metallic plate $A$. The photo electrons are accelerated by a potential difference $10\text{V}$. These electrons strike another metal plate $B$ from which electromagnetic radiations are emitted. The minimum wavelength of the emitted photons is $100\text{nm}$. The work function of the metal $A$ is $x\text{eV}$, then the value of $x+2.2$ is (use $hc=12400\text{eV}\stackrel{\circ }{A}$)

Answer 1

${\lambda }_{\text{min}}\Rightarrow f_{\text{max}}\Rightarrow E_{\text{max}}$.
Thus the maximum velocity photoelectron loses all its energy to radiate ${\lambda }_{\text{min}}$
$\Rightarrow \frac{1}{2}m{V}_{\text{max}}^2+e\left(10\right)=\frac{hc}{{\lambda }_{\text{min}}}$

From Einstein's Photo-Electric equation (All terms in $eV$)
$\frac{hc}{\lambda e}=x+\frac{1}{2}m{V}_{\text{max}}^2$
$\Rightarrow \frac{hc}{e\lambda }=x+\frac{hc}{e{\lambda }_{\text{min}}}-10$
$\Rightarrow x=\frac{hc}{e}[\frac{1}{\lambda }-\frac{1}{{\lambda }_{\text{min}}}]+10$
$x=12400\times {10}^{-10}[\frac{{10}^9}{200}-\frac{{10}^9}{100}]+10$
$x=-6.2+10$
$x=3.8$
$\therefore x+2.2=3.8+2.2=6$