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11.Dual Nature of Radiation and matter
easy
Find the number of photons 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
$\frac{25}{2} \times 10^{19} , 0.4\,amp$
B
$\frac{25}{4} \times 10^{19} , 6.2\,amp$
C
$\frac{25}{2} \times 10^{19} , 0.8\,amp$
D
None of these
Solution
(a)
$n =\frac{P}{\frac{h c}{\lambda}}=\frac{P \lambda}{h c}=\frac{25 \times 6600 \times 10^{-10}}{6.64 \times 10^{-34} \times 3 \times 10^9}$
$=8.28 \times 10^{19}=\frac{25}{3} \times 10^{19}$
$3 \%$ of emitted photons are producing current
$I=\frac{3}{100} \times n e=\frac{3}{100} \times \frac{25}{3} \times 10^{19} \times 1.6 \times 10^{-19}=04\,A$
Standard 12
Physics
