Question

In a photoelectric effect set-up a point source of light of power $3.2\times {10}^{-3}W$ emits monoenergetic photons of energy $5.0eV$. The source is located at a distance of $0.8m$ from the centre of a stationary metallic sphere of work function $3.0eV$ and of radius $8.0\times {10}^{-3}m$. The efficiency of photoelectron emission is one-for every ${10}^6$ incident photons. Assume that the sphere is isolated and initially natural and that photoelectron are instantly swept away after emission.
Calculator the number of photoelectron emitted per second.
Find the ratio of the wavelength of incident light to the de-Broglie wavelength of the fastest photoelectron emitted.
It is observed that the photoelectron emission stop at a certain time $t$ after the light source is switch on. Why?
Evaluate the time $t$.

Answer 1

Energy of emitted photons
$E_1=0.5eV=8\times {10}^{-19}$
Energy emitted per second by the source
$E_2=3.2\times {10}^{-3}J$
No. of photons emitted per second $=\frac{E_2}{E_1}=4\times {10}^{15}$
No. of photons incident on the sphere per second
$n=\frac{4\times {10}^{15}}{4\pi {\left(0.8\right)}^2}\times \pi {(8\times {10}^{-3})}^2\approx {10}^{11}$ per second
No, of photoelectron emitted per second
$n^{\prime }=\frac{{10}^{11}}{{10}^6}={10}^5$ per second
Max. $K.E.$ of photo electron
$k_{max}=$ Energy of incident photons work function $=2eV=3.2\times {10}^{-19}J$
de-Broglie wavelength of these photoelectrons
${\lambda }_1=\frac{h}{\sqrt{2k_{max}{m}_e}}=8.68A$
wavelength of incident light
${\lambda }_2=\frac{12375}{5}=2475A$
The desired ratio is ${\lambda }_1/{\lambda }_2$
Emission of photoelectrons is stopped when its potential is equal to the stopping potential required of the fastest moving electrons
$k_{max}=2eV$
Stopping potential $v_0=2volt$
$\Rightarrow 2=kq/r$
$\Rightarrow q=1.78\times {10}^{-12}C$
$\Rightarrow t=\frac{q}{{10}^5\times 1.6\times {10}^{-19}}$
$t\approx 111$ seconds.