Question

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 $h=6.6 \times 10^{-34} \,Js$ )

• A. $48 \times 10^{19}$
• B. $48 \times 10^{17}$
• C. $8 \times 10^{19}$
• D. $24 \times 10^{17}$

Answer 1

Answer: (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 $=\frac{3}{100} \times \frac{24}{h c} \times \lambda$