Question

The light of radiation $300nm$ falls on a photocell operating in the saturation mode. The spectral sensitivity is $4.8mA/W.$ The yield of photo electrons (i.e. number of electrons produced per photon) is

• A. $0.04$
• B. $0.02$
• C. $0.03$
• D. $0.2$

Answer 1

The correct option is D $0.2$

energy per photon per second = $\frac{hC}{\lambda }$

= $\frac{6.663\times {10}^{-34}\times 3\times {10}^8}{3\times {10}^{-11}}$

= $6.6\times {10}^{-19}J/s$
= $6.6\times {10}^{-19}W$
$1W$ gives $4.8mA$
then $6.6\times {10}^{-15}$ W gives = $(4.8\times {10}^{-3}\times 6.6\times {10}^{-19})A$
$=31.7\times {10}^{-21}A$
assuming n no. of electrons per second gives $31.7\times {10}^{-8}A$ of current, then

$I=\frac{\Delta Q}{\Delta t}$

$31.7\times {10}^{-21}=\frac{n\times 1.6\times {10}^{-19}}{1}$

$n=0.2$