Question

One mili watt of light of wavelength $4560\, \mathring{A} $ is incident on a cesium surface of work function $1.9 \,eV$. Given that quantum of efficiency of photoelectric emission is $0.5\%$, Planck constant $ h=6.62\times 10^{-34}J-s, $ velocity of light $ c=3\times 10^{8} m/s, $ the photoelectric current liberated is :

• A. $ 1.84\times 10^{-6}A $
• B. $ 1.84\times 10^{-7}A $
• C. $ 1.84\times 10^{-5}A $
• D. $ 1.84\times 10^{-4}A $

Answer 1

Answer: (A)

From Planck's law, the energy of incident photon is
$E=n h v=\frac{n h c}{\lambda}$
where $n$ is number of photons.
$\Rightarrow n=\frac{E}{h c / \lambda}$
Given, $E=1\, mW =10^{-3} W $,
$ h=6.62 \times 10^{-34} J -s$,
$c=3 \times 10^{8} J-s$,
$ \lambda=4560 \, \mathring{A}=4560 \times 10^{-10} m$
$n=\frac{10^{-3} \times 4.56 \times 10^{-7}}{6.62 \times 10^{-34} \times 3 \times 10^{8}}=2.29 \times 10^{15}$
Given, quantum efficiency $=0.5 \%$.
Hence, number of electrons liberated from surface is
$n=n \frac{0.5}{100}=1.15 \times 10^{13} \therefore $
$l=n e=1.15 \times 10^{13} \times 1.6 \times 10^{-19} i$
$=1.84 \times 10^{-6} A$