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

Answer 1

Answer: (A)

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