Question

A mercury arc lamp provides $0.1\text{Watt}$ of ultra-violet radiation at a wavelength of $\lambda =2537\stackrel{\circ }{A}$ only. The photo tube (cathode of photo electric device) consists of potassium and has an effective area of $4c{m}^2$. The cathode is located at a distance of $1\text{m}$ from the radiation source. The work function for potassium is ${\varphi }_0=2.2eV$.

According to classical theory, the radiation from arc lamp spreads out uniformly in space as spherical wave. What time of exposure to the radiation should be required for a potassium atom $\left(\text{radius}2\stackrel{\circ }{A}\right)$ in the cathode to accumulate sufficient energy to eject a photoelectron?

• A. $352\text{s}$
• B. $176\text{s}$
• C. $704\text{s}$
• D. No time lag

Answer 1

The correct option is A $352\text{s}$

Given that
$P=0.1\text{W}$
Consider an imaginary sphere at of radius $1m$ over the source because it is producing spherical wave and let the potassium atom be touching the sphere.
So Power effective at $1m$ transfered to potassium atom will be
$P_{eff}=\frac{P}{4\pi r_s^2}\times \pi r_A^2\text{where}\pi r_A^2\text{is the area of potassium atom}$
Given that work function $\varphi =2.2\text{eV}=2.2\times 1.6\times {10}^{-19}\text{J}$
So
$\Delta t=\frac{\varphi }{P_{eff}}=\frac{2.2\times 1.6\times {10}^{-19}}{\frac{P}{4\pi r_s^2}\times \pi r_A^2}$
Putting values and solving we get
$\Delta t=352s$