Question

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

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

Answer 1

Answer: (C)

Let ${\varphi }_0$ be the work function of the surface of the material.
Then, According to Einstein's photoelectric equation, the maximum kinetic energy of the emitted photoelectrons in the first case is
$K_{{\max }_1}=\frac{hc}{\lambda }-{\varphi }_0$
and that in the second ease is
$K_{{\max }_2}=\frac{hc}{\lambda }-{\varphi }_0=\frac{2hc}{\lambda }-{\varphi }_0$
But $K_{\max 2}=3K_{\max 1}($ given $)$
$\therefore \frac{2hc}{\lambda }-{\varphi }_0=3\left(\frac{hc}{\lambda }-{\varphi }_0\right)$
$\frac{2hc}{\lambda }-{\varphi }_0=\frac{3hc}{\lambda }-3{\varphi }_0$
$3{\varphi }_0-{\varphi }_0=\frac{3hc}{\lambda }-\frac{2hc}{\lambda }$
$2{\varphi }_0=\frac{hc}{\lambda }$
or ${\varphi }_0=\frac{hc}{2\lambda }$