Question

When a metal surface is illuminated by a monochromatic light of wavelength ${}^{\prime }{\lambda }^{\prime }$, then the potential difference required to stop the ejection of electrons is 3v. When the same surface is illuminated by the light of wavelength 2$\lambda$, then the potential difference required to stop the ejection of electrons is v V. Then for photoelectric effect work function is $\frac{hc}{x}$. Find x

• A. 6$\lambda$
• B. 4$\lambda$/3
• C. 4$\lambda$
• D. 8$\lambda$

Answer 1

The correct option is C 4$\lambda$

$3v\ast e=\frac{hc}{\lambda }-\varphi$---------(1)
$v\ast e=\frac{hc}{2\lambda }-\varphi$---------(2)
substituting value of v*e from 2 into 1
$3(\frac{hc}{2\lambda }-\varphi )=\frac{hc}{\lambda }-\varphi$

$\frac{3hc}{2\lambda }-3\varphi =\frac{hc}{\lambda }-\varphi$

$\frac{hc}{2\lambda }=2\varphi$

$\varphi =\frac{hc}{4\lambda }$

So, the answer is (C).