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

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