Question

In fig, electromagnetic radiations of wavelength $200nm$ are incident on a metallic plate A. The photoelectrons are accelerated by a potential difference of $10V$.These electrons strike another metal plate B from which electromagnetic radiations are emitted. The minimum wavelength of emitted photons is $100nm$. If the work function of metal A is found to be $(x\times {10}^{-1})eV$, then find the value of $x$. (Given $hc=1240eVnm)$

Answer 1

Let us first find the initial energy of the electron just after the light strikes the plate :
$E_i$ = $\frac{hC}{\lambda }$ - $\varphi$ = $\frac{1240}{200}-\varphi$ = $6.2-\varphi$
After this it gains energy from the battery. Now the total energy becomes :
$E_f$ = $E_i$ + $10$ = $16.2-\varphi$
This energy is released as light from the other plate .
$\frac{hC}{{\lambda }_2}$ = $16.2-\varphi$
$\frac{1240}{100}$ = $16.2-\varphi$
$\varphi =16.2-12.4=3.8=38\times {10}^{-1}$

Hence $x=38$