Question

When a light of photons of energy $4.2eV$ is incident on a metallic sphere of radius $10cm$ and work function $2.4eV$, photoelectrons are emitted. The number of photoelectrons liberated before the emission is stopped, is ($e=1.6\times {10}^{-19}C$ and $\frac{1}{4\pi {\epsilon }_0}=9\times {10}^{9}N{m}^2/C^2$):

• A. $6.25\times {10}^8$
• B. $1.25\times {10}^9$
• C. $1.25\times {10}^8$
• D. $6.25\times {10}^{18}$

Answer 1

The correct option is C $1.25\times {10}^8$

Incident photons releases electrons from the sphere and hence the sphere gains a positive charge. A single photon releases only a single electron. Assuming all the energy is transferred to the sphere as electrostatic potential energy after overcoming the work function, the potential energy gained by the sphere equals the total energy of photons.

Let number of incident photons be n.

Then, potential energy of charged sphere is given by:

$E_1=\frac{Q^2}{4\pi {ϵ}_o{R}^2}=\frac{(ne{)}^2}{4\pi {ϵ}_{o}R}$

Total energy of incident photons imparted to the sphere after overcoming the work function is:

$E_2=n(E_p-\varphi )$

$E_1=E_2$

$\frac{n^2{e}^2}{4\pi {ϵ}_{o}R}=n(E_p-\varphi )$

$n=\frac{4\pi {ϵ}_{o}R(E_p-\varphi )}{e^2}$

$n=\frac{0.1\times (4.2-2.4)\times 1.6\times {10}^{-19}}{9\times {10}^9\times (1.6\times {10}^{-19}{)}^2}$

$n=1.25\times {10}^8$