Question

Ejection of the photoelectron from metal in the photoelectric effect experiment can be stopped by applying $0.5V$ when the radiation of $250nm$ 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

$hv=h{v}_0+(K.E.{)}_{\max }$

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

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

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

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

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