Question

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

• A. $\frac{25}{3}\times {10}^{19}J,0.4A$
• B. $\frac{25}{4}\times {10}^{19}J,6.2A$
• C. $\frac{25}{2}\times {10}^{19}J,0.8A$
• D. $\frac{25}{2}\times {10}^{19}J,4A$

Answer 1

The correct option is A $\frac{25}{3}\times {10}^{19}J,0.4A$

$P_{in}=25W,\lambda =6600\stackrel{\circ }{A}=6600\times {10}^{-10}m$
$nhv=P$
$\Rightarrow$ 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
$\therefore I=\frac{3}{100}\times ne=\frac{3}{100}\times \frac{25}{3}\times {10}^{19}\times 1.6\times {10}^{-19}=0.4A$.