Question

In a photoelectric effect set-up a point of light of power $3.2\times10^{-3}$ W emits monoenergetic photons of energy 5.0 eV. The source is located at a distance of 0.8 m from the centre of a stationary metallic sphere of work function 3.0 eV and of radius $8.0\times10^{-3}m$. The efficiency of photoelectrons emission is one for every $10^6$ incident photons. Assume that the sphere is isolated and initially neutral and that photoelectrons are instantly swept away after emission
(a) Calculate the number of photoelectrons emitted per second
(b) Find the ratio of the wavelength of incident light to the de-Broglie wavelength of the fastest photoelectrons emitted. (c) It is observed that the photoelectrons emission stops at a certain time t after the light source is switched on why ? (d) Evaluate the time t.

Answer 1

(a) Energy of emitted photons
$E_1 = 5.0 eV= 5.0\times1.6\times10^{-I9}J$
$=8.0\times10^{-19} J$
Power of the point source is $3.2\times10^{-3}$ W or $3.2\times10^{-3} J/s.$
Therefore, energy emitted per second,
$E_2 = 3.2\times10^{-3} J$
Hence, number of photons emitted per second $n_1=\frac{E_2}{E_1}$
or $\frac{3.2\times10^{-3}}{8.0\times10^{-19}}$
$n_1=4.0\times10^{15}photon/s$
Number of photons incident on unit area at a distance of 0.8 m from the source S will be $n_2=\frac{n_1}{4\pi(0.8)^2}=\frac{4.0\times10^{15}}{4\pi(0.64)}$
$\approx5\times10^{14}photon/s-m^2$
The area of metallic sphere over which photons will fall is
$A=\pi r^2=\pi(8\times10^{-3})^2m^2\approx2.01\times10^{-4}m^2$
The area of metallic sphere over which photons will fall is
$n_3=n_2 A=(5.0\times10^{14}\times2.01\times10^{-4})=10^{11}/s$
But since, one photoelectron is emitted for every $10^6$ photons, hence number of photoelectrons emitted per second.
$n=\frac{n_3}{10^6}=frac{10^{11}}{10^6}=10^5/s or n=10^5/s$
(b) Maximum kinetic energy of photoelectrons
$K_{max}$= Energy of incident photons - work function
= (5.0- 3.0) eV = 20 eV
$=2.0\times1.6\times10^{-19}J$
$K_{max} =3.2 x 10^{-19}J$
The de-Broglie wavelength of these photoelectrons will be
$\lambda_1=\frac{h}{p}=\frac{h}{\sqrt{2K_{max}m}}$
Here h = Planck's constant and m = mass of electron
$\lambda_1=\frac{6.63\times10^{-34}}{\sqrt{2\times3.2\times10^{-19}\times9.1\times10^{-31}}}$
$=8.68\times10^{-10}=8.68\mathring{A}$
Wavelength of incident light $\lambda_2(in \mathring{A})=\frac{12375}{E_1}(in eV)$
or $\lambda_2=\frac{12375}{5}=2475\mathring{A}$
Therefore, the desired ratio is
$\frac{\lambda_2}{\lambda_1}=\frac{2475}{8.68}=285.1$
(c) As soon as electrons are emitted from the metal sphere, it gets positively charged and acquires positive potential. The positive potential gradually increases as more and more photoelectrons are emitted from its surface. Emission of photoelectrons is stopped when its potential is equal to the stopping potential required for fastest moving electrons.
(d) As discussed in part (c) emission of photoelectrons is stopped when potential on the metal sphere is equal to the stopping potential of fastest moving electrons. Since,
$K_{max}=2.0eV$
Therefore, stopping potential $V_0 = 2 V.$ Let q be the charge required for the potential on the sphere to be equal to stopping potential or 2 V. Then
$2=\frac{1}{4\pi\varepsilon_0}.\frac{q}{r}=(9.0\times10^9)\frac{q}{8.0\times10^{-3}}$
$\therefore q=1.78\times10^{-12}C$
Photoelectrons emitted per second = $10^5$ [Part (a)]
or charge emitted per second $=(1.6\times$10^{-19})\times10^5C
$1.6\times10^{-14}C$
Therefore, time required to acquire the charge q will be
$t=\frac{q}{1.6\times10^{-14}}s=\frac{1.78\times10^{-12}}{1.6\times10^{-14}}s$ or $t\approx111s$