Question

Choose the correct answer from the alternatives given.
A parallel beam of monochromatic radiation of cross-section area A $(<\pi a^2)$, intensity I and frequency $\upsilon$ is incident on a solid conducting sphere of work function ${\varphi }_o(hv>{\varphi }_0)$ and radius a. The sphere is grounded by a conducting wire. Assume that for each incident photon, one photoelectron is ejected. Just after this radiation is incident on initially unchanged sphere the current through the conducting wire is

• A. $\frac{IAe}{hv}$
• B. $\frac{IAe}{2hv}$
• C. $\frac{2IAe}{hv}$
• D. $\frac{2}{3}\frac{IAe}{hv}$

Answer 1

Answer: (A)

$\frac{IAe}{hv}$

Energy of each incident photon = $h\upsilon$
Total energy incident per second on sphere = IA
$\therefore$ No. of photons incident on sphere per second = $\frac{IA}{h\upsilon }$
$\therefore$ No. of photolectrons ejected per second = $\frac{IA}{h\upsilon }$
Current through the conducting wire
= charge flowing per second through wire = $\frac{IAe}{h\upsilon }$