Question

The variation of photo-current with collector potential for different frequencies of incident radiation ${\nu }_1,{\nu }_2$ and ${\nu }_3$ is as shown in the graph. Then:

• A. ${\nu }_1={\nu }_2={\nu }_3$
• B. ${\nu }_1>{\nu }_2>{\nu }_3$
• C. ${\nu }_1<{\nu }_2<{\nu }_3$
• D. ${\nu }_3=\frac{{\nu }_1+{\nu }_2}{2}$

Answer 1

The correct option is C ${\nu }_1<{\nu }_2<{\nu }_3$

The points $V_{O1},V_{O2}\&V_{O3}$ on the graph corresponds to stopping potential with $V_{O3}>V_{O2}>V_{O1}$.

Minimum energy of photon required for emission is proportional to the stopping potential.

Hence, $E_3>E_2>E_1$

Now, Energy of photon is given by $E=h\nu$

$⟹{\nu }_3>{\nu }_2>{\nu }_1$