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 neutral, and that photoelectron are instantly swept away after emission. If the time $t$ is given by $1.11\times {10}^{-k}$. Then $k$ will be given by

Answer 1

solution:

let $E_1=0.5eV=8\times {10}^{-19}J$ and

$E_2=3.2\times {10}^{-3}J$

No. of photons emitted per second = $\frac{E_2}{E_1}=4\times {10}^{15}$No. of photons incident on the sphere per second

$n=\frac{4\times {10}^{15}}{4\pi (0.8{)}^2}\times \pi (8\times {10}^{-3}{)}^2\approx {10}^{11}$ per second

No, of photoelectron emitted per second

$n^{\prime }={10}^{11}{10}^6={10}^5$ 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 $\frac{{\lambda }_1}{{\lambda }_2}$

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 $v_0=2volt$

$\to 2=kq/r$

$\to q=1.78\times {10}^{-12}C$

$\to t=\frac{q}{{10}^5\times 1.6\times {10}^{-19}}$

$t\approx 111seconds=1.11$ x ${10}^2$

hence k =2

hence the correct answer is 2