Question

Light of wavelength $3500\,\mathring{A}$ is incident on two metals $A$ and $B$, $A$ of work function $4.2 \,eV$ and $B$ of work function $1.19 \,eV$ respectively. The photoelectrons will be emitted by :

• A. metal A
• B. metal B
• C. both A and B
• D. neither metal A nor metal B

Answer 1

Answer: (B)

The minimum energy required for the emission of photoelectrons from a metal is called the work function of that metal.
From Planck's law energy $(E)$ of incident light is
$ E=\frac{hc}{\lambda } $ where $ \lambda $ is wavelength.
Given, $ \lambda =3500\, \mathring{A}=3500\times 10^{-10} m$,
$ h=6.64\times 10^{-34} J-s, $
$ c=3\times 10^{8}m/s $
$ E=\frac{6.64\times 10^{-34}\times 3\times 10^{8}}{3500\times 10^{-10}} $
$ =5.69\times 10^{-19},J $
Also, $ 1\,eV=1.6\times 10^{-19}J $
$ \therefore $ $ E=\frac{5.69\times 10^{-19}}{1.6\times 10^{-19}} eV=3.55\,eV $
Hence, only $B$ will emit photoelectrons as incident energy is greater than work function.