Question

A photoelectric surface is illuminated successively by monochromatic light of wavelength $\lambda$ and $\frac{\lambda }{2}$. 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. $\frac{hc}{3\lambda }$
• B. $\frac{hc}{2\lambda }$
• C. $\frac{hc}{\lambda }$
• D. $\frac{hc}{4\lambda }$

Answer 1

The correct option is B $\frac{hc}{2\lambda }$

In first case, $E_0=\frac{hc}{\lambda }-\varphi ......\left(1\right)$

and in 2nd case, $3E_0=\frac{hc}{\lambda /2}-\varphi .....\left(2\right)$

$\left(2\right)/\left(1\right)\Rightarrow \frac{3}{1}=\frac{\frac{2hc}{\lambda }-\varphi }{\frac{hc}{\lambda }-\varphi }$

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

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