Question 21: Dual Nature of Matter and Radiation

Question

When a certain metal surface is illuminated with light of frequency $
u$, the stopping potential for photoelectric current is $V_0$. When the same surface is illuminated by light of frequency $\frac{
u}{2}$, the stopping potential is $\frac{V_0}{4}$. The threshold frequency for photoelectric emission is

Accepted Answer

We know that,
$e V_0=h 
u-\phi_0$
where, $V_0=$ stopping potential
and $\quad 
u=$ frequency of light.
Case I $e V_0=h 
u-\phi_0$ .... (i)
Case II $\frac{e V_0}{4}=\frac{h 
u}{2}-\phi_0$
$\Rightarrow \quad e V_0=2 h 
u-4 \phi_0$ .... (ii)
From Eqs. (i) an i (ii), we get
$h 
u-\phi_0 =2 h 
u-4 \phi_0$
$ \Rightarrow h
u  = 3{\phi _0} \Rightarrow h
u  = 3h{
u _0}\quad$ ($\because$ ${\phi _0} = h{
u _0}$)
$ \Rightarrow {
u _0} = {{h
u } \over {3h}} \Rightarrow {
u _0} = {
u  \over 3}$

Answer

$\frac{
u}{6}$

Answer

$\frac{
u}{3}$

Answer

$\frac{2
u}{3}$

Answer

$\frac{4
u}{3}$

Dual Nature of Matter and Radiation

When a certain metal surface is illuminated with light of frequency \nu, the stopping potential for photoelectric current is V_0. When the same surface is illuminated by light of frequency \frac{\nu}{2}, the stopping potential is \frac{V_0}{4}. The threshold frequency for photoelectric emission is

Choose the correct answer: