11.Dual Nature of Radiation and matter
easy

Find the number of electrons emitted per second by a $24 \,W$ source of monochromatic light of wavelength $6600 \mathring A$, assuming $3 \%$ efficiency for photoelectric effect (take $h=6.6 \times 10^{-34} \,Js$ ) $S$

A

$48 \times 10^{19}$

B

$48 \times 10^{17}$

C

$8 \times 10^{19}$

D

$24 \times 10^{17}$

Solution

(d)

Energy per photon $=h f$ or $\frac{h c}{\lambda}$

Number of photon's per second $=\frac{24}{h f}$

Number of electrons emitted $=\frac{3}{100} \times \frac{24}{h c} \times \lambda$

Standard 12
Physics

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