Solution:
Solution
Incident energy
$$E = 10.6 \, eV = 10.6 \times (1.6 \times 10^{-19}) \, J = 16.96 \times 10^{-19} \, J$$
Given : $$\dfrac{Energy \, incident}{area \times time} = 2 W/m^2$$
$$\therefore \dfrac{Number \, of \, incident \, photons}{area \times time} = \dfrac{2}{16.96 \times 10^{-19}} \,\, \, \, \, = \, 1.18 \times 10^{18}$$
$$\therefore \dfrac{Incident \, photons}{time} = (1.18 \times 10^8) \times area$$
$$ = 1.18 \times 10^{18} \times (1.0 \times 10^{-4}) = 1.18 \times 10^{14}$$
$$\therefore \dfrac{Number \, of \, photoelectrons}{time} = \left(\dfrac{0.53}{100}\right) \times (1.18 \times 10^{14})$$
or $$n = 6.25 \times 10^{11}$$