Question

$K_{1}and{K}_2$ are the maximum kinetic energies of the photo electrons emitted when light of wave lengths ${\lambda }_{1}and{\lambda }_2$ respectively are incident on a metallic surface. If ${\lambda }_1=3{\lambda }_2$ then

• A. (a) $K_1>\frac{K_2}{3}$
• B. (b) $K_1<\frac{K_2}{3}$
• C. (c) $K_1>3K_2$
• D. (d) $K_2=3K_1$

Answer 1

The correct option is D (d) $K_2=3K_1$

For photoelectrons having maximum kinetic energy

$K=\frac{hc}{\lambda }\therefore K_1=\frac{hc}{{\lambda }_1}and{K}_2=\frac{hc}{{\lambda }_2}Now,{\lambda }_1={3\lambda }_2\Rightarrow K_1=\frac{hc}{{3\lambda }_2}{K}_1=\frac{K_2}{3}\Rightarrow K_2=3K_1\left(d\right)$