Question

When a metallic surface is illuminated with a monochromatic light of wavelength $\lambda$, the stopping potential is $5V_0$. When the same surface is illuminated with light of wavelength 3$\lambda$, the stopping potential is $V_0$. Then the work function of the metallic surface is:

• A. $\frac{hc}{6\lambda }$
• B. $\frac{hc}{5\lambda }$
• C. $\frac{hc}{4\lambda }$
• D. $\frac{2hc}{4\lambda }$

Answer 1

Answer: (A)

$\frac{hc}{6\lambda }$

From Einstein photoelectric equation we get,when wavelength $=\lambda$ and stopping potential $=5V_0$

$\frac{hc}{\lambda }=e\left(5V_0\right)+W$-------------(A)

where W=work function

similarly when wavelength$=3\lambda$ and stopping potential is $V_0$

$\frac{hc}{3\lambda }=e{V}_0+W$

$\frac{hc}{3\lambda }-W=e{V}_0$-----------(B)

Putting B in A we get,

$\frac{hc}{\lambda }=\frac{5hc}{3\lambda }-5W+W$

$4W=\frac{2hc}{3\lambda }$

$W=\frac{hc}{6\lambda }$