Question

Find the number of electrons emitted per second by a $24W$ source of monochromatic light of wavelength $6600A^o$, 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

The correct option is D $24\times {10}^{17}$

Energy per second = $24J$

Energy of a photon =$h\nu$

=$6.626\times {10}^{-34}\times \frac{3\times {10}^8}{6600\times {10}^{-10}}$

= $3\times {10}^{-19}J$

$\therefore$ Number of photons per second incoming = $\frac{24}{3\times {10}^{-19}}=8\times {10}^{19}$

Considering $3\%$ efficiency, number of electron emmited = $\frac{3}{100}\times 8\times {10}^{19}=24\times {10}^{17}$