Question

A photoelectric surface is illuminated successively by monochromatic light of wavelengths $\lambda$ and $\lambda / 2$. If the maximum kinetic energy of the emitted photoelectrons in the second case is 3 times than in the first case, the work function of the surface of the material is
( $h=$ Planck's constant, $c=$ speed of light)

• A. $\frac{h c}{2 \lambda}$
• B. $\frac{h c}{\lambda}$
• C. $\frac{2 h c}{\lambda}$
• D. $\frac{h c}{3 \lambda}$

Answer 1

Answer: (A)

According to Einstein photoelectric equation
$E=K_{\max }+\phi$
where, $K_{\max }$ is maximum kinetic energy of emitted electron and $\phi$ is work function of an electron
$K_{\max }=E-\phi=h v-\phi $
$K_{\max }=\frac{h c}{\lambda}-\phi$
Similarly, in second case, maximum kinetic energy of emitted electron is $3$ times that in first case, we get
$3 K_{\max }=\frac{h c}{\lambda / 2}-\phi$
Solving Eqs. (i) and (ii), we get work function of an emitted electron from a metal surface
$\phi=\frac{h c}{2 \lambda}$