Question

Light of wavelength $3500\mathring{A}$ is incident on two metals $A$ and $B$, $A$ of work function $4.2eV$ and $B$ of work function $1.19eV$ 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, $1eV=1.6\times 10^{-19}J$
$\therefore$ $E=\frac{5.69\times 10^{-19}}{1.6\times 10^{-19}}eV=3.55eV$
Hence, only $B$ will emit photoelectrons as incident energy is greater than work function.