Question

A mercury arc lamp provides $0.1\text{W}$ 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 ${\varphi }_0=2.2eV$.

To what saturation current does the flux of photons at the cathode corresponds, if the photo conversion efficiency is $5\%$?

• A. $32.5nA$
• B. $10.15nA$
• C. $65nA$
• D. $3.25nA$

Answer 1

The correct option is A $32.5nA$

Power received by the plate
$P_{eff}=\frac{P}{4\pi (1{)}^2}\times 4\times {10}^{-4}$
So number of photons emiited per unit time can be given as
$\frac{\Delta n_p}{\Delta t}=\frac{P_{eff}}{E_p}=\frac{0.1}{4\pi (1{)}^2}\times \frac{4\times {10}^{-4}\lambda }{hc}$
On solving and putting values
$\frac{\Delta n_p}{\Delta t}=4.06\times {10}^{12}$
So number of photo-electrons as per efficiency 5% is given as
$\frac{\Delta n_q}{\Delta t}=\frac{5}{100}\frac{\Delta n_p}{\Delta t}$
So by multiplying number of photoelectrons with charge of an electron we can get photocurrent.
$i=\frac{\Delta n_q}{\Delta t}e=e\frac{5}{100}\frac{\Delta n_p}{\Delta t}$
On solving
$i=32.5nA$