Question

Find the number of photon emitted per second by a $25$ watt source of monochromatic light of wavelength $6600\mathring{A}$. What is the photoelectric current assuming $3\%$ efficiency for photoelectric effect ?

• A. $\frac{25}{3}\times 10^{19}J,0.4amp$
• B. $\frac{25}{4}\times 10^{19}J,6.2amp$
• C. $\frac{25}{2}\times 10^{19}J,0.8amp$
• D. None of there

Answer 1

Answer: (A)

$P_{\text{in }}=25W,\lambda =6600\mathring{A}=6600\times 10^{-10}m$
$nhv=P$
$\Rightarrow$ Number of photons emitted / sec,
$n=\frac{P}{\frac{hc}{\lambda }}=\frac{P\lambda }{hc}=\frac{25\times 6600\times 10^{-10}}{6.64\times 10^{-34}\times 3\times 10^8}$
$=8.28\times 10^{19}=\frac{25}{3}\times 10^{19}$
$3\%$ of emitted photons are producing current
$\therefore I=\frac{3}{100}\times ne$
$=\frac{3}{100}\times \frac{25}{3}\times 10^{19}\times 1.6\times 10^{-19}=0.4A$.