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. $\frac{hc}{2\lambda }$
• B. $\frac{hc}{\lambda }$
• C. $\frac{2hc}{\lambda }$
• D. $\frac{hc}{3\lambda }$

Answer 1

Answer: (A)

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