Question

Ejection of the photoelectron from metal in the photoelectric effect experiment can be stopped by applying $0.5\, V$ when the radiation of $250\, nm$ is used. The work function of the metal is :

• A. 4 eV
• B. 4.5 eV
• C. 5 eV
• D. 5.5 eV

Answer 1

Answer: (B)

From photoelectric experiment, we have

$h v=h v_{0}+( K.E. )_{\max }$

$\frac{h v}{\lambda}=W+( K . E .)_{\max }$

where $\frac{h v}{\lambda}$ is the energy of incident radiation, $W$ is the work function and $( K . E .)_{\max }$ is the kinetic energy of the ejected electrons.

$\frac{h v}{\lambda} =W+( K.E. )_{\operatorname{max}}$

$\frac{6.6 \times 10^{-34} \times 3 \times 10^{8}}{256.7 \times 10^{9} \times 1.6 \times 10^{-19}}$ $=W+0.5$ (As $1\, V =1\, eV$ )

$4.95\, eV =W+0.5\, eV \Rightarrow W=4.45\, eV$