When a light of photons of energy 4.2 eV is incident on a metallic sphere of radius 10 cm and work function 2.4 eV, photoelectrons are emitted. The number of photoelectrons liberated before the emission is stopped, is ($e=1.6×10^{-19}C$ and $\frac{1}{4πε_0}=9×10^9Nm^2C^{-2}$)

$1.25×10^8$

Here,

Radius of the sphere, r = 10 cm = $10×10^{-2}m$

Work function, $\phi_0 = 2.4eV$

Energy of a photon, E = hv = 4.2eV

According to Einstein’s photoelectric equation

$hv=\phi_0+eV_s$

$4.2eV=2.4eV+eV_s$

$eV_s=4.2eV-2.4eV=1.8eV$

$∴V_s=1.8V$

The sphere will stop emitting photoelectrons, when the potential on its surface becomes 1.8 V.

Let N be the number of photoelectrons emitted from the sphere. Then,

Charge on the sphere, Q = Ne

$V_s=\frac{1}{4πε_0}\frac{Q}{r}=\frac{1}{4πε_0}\frac{Ne}{r}$

$N=\frac{V_s×r}{\frac{1}{4πε_0}×e}=\frac{1.8×r}{\frac{1}{4πε_0}×e}=\frac{1.8×10×10^{-2}}{9×10^9×1.6×10^{-19}}=\frac{18}{19}×\frac{1}{9}×10^9=1.25×10^8$