Question

One mili watt of light of wavelength $4560\mathring{A}$ is incident on a cesium surface of work function $1.9eV$. 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
$E=nhv=\frac{nhc}{\lambda }$
where $n$ is number of photons.
$\Rightarrow n=\frac{E}{hc/\lambda }$
Given, $E=1mW=10^{-3}W$,
$h=6.62\times 10^{-34}J-s$,
$c=3\times 10^{8}J-s$,
$\lambda =4560\mathring{A}=4560\times 10^{-10}m$
$n=\frac{10^{-3}\times 4.56\times 10^{-7}}{6.62\times 10^{-34}\times 3\times 10^8}=2.29\times 10^{15}$
Given, quantum efficiency $=0.5\%$.
Hence, number of electrons liberated from surface is
$n=n\frac{0.5}{100}=1.15\times 10^{13}\therefore$
$l=ne=1.15\times 10^{13}\times 1.6\times 10^{-19}i$
$=1.84\times 10^{-6}A$