Question

A metallic surface of work function hv is illuminated by a radiation beam of frequency 5v. Stopping potential observed is X. What will be the stopping potential if the surface is illuminated by radiations of the of 7v frequency?

Answer 1

By Einstein's photolectric equation we have:

$h{v}_{incident}=\varphi +\frac{1}{2}m{V}^2$ (where $\varphi =$ work function and $v=$ velocity of electron having max $K.E$)

Also stopping potential $=$ Max K.E (For stopping the electron having max $K.E$)

$\Rightarrow h\left(5v\right)=hv+X$ ($X=$ stopping potential for $1^{st}$ care)

$\Rightarrow X=4hv\to \left(1\right)$

When incident light of frequency $7v$ strikes then let stopping potential be $X_2$

$\Rightarrow h\left(7v\right)=hv+X_2$

$\Rightarrow X_2=6hv\to \left(2\right)$

Dividing equation 2 by 1 we have:

$\frac{X_2}{X}=\frac{6}{4}=\frac{3}{2}$

$\Rightarrow X_2=\frac{2}{3}X\to$ New stopping potential