One milliwatt of light of wavelength 4560 ma | vedclass

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Question

(A) The power of incident light $P = 1 \ mW = 10^{-3} \ W$. The wavelength $\lambda = 4560 \ \mathring{A} = 4560 	imes 10^{-10} \ m$. First,calculate

Vedclass · Accepted Answer

$1.856 	imes 10^{-6} \ A$

Vedclass · Answer

(A) The power of incident light $P = 1 \ mW = 10^{-3} \ W$. The wavelength $\lambda = 4560 \ \mathring{A} = 4560 	imes 10^{-10} \ m$. First,calculate the number of photons incident per second $(n_p)$: $n_p = \frac{P}{E_{photon}} = \frac{P \lambda}{hc} = \frac{10^{-3} 	imes 4560 	imes 10^{-10}}{6.62 	imes 10^{-34} 	imes 3 	imes 10^8} \approx 2.295 	imes 10^{15} 	ext{ photons/s}$. The number of photoelectrons emitted per second $(n_e)$ is given by the quantum efficiency $\eta = 0.5\% = 0.005$: $n_e = \eta 	imes n_p = 0.005 	imes 2.295 	imes 10^{15} \approx 1.1475 	imes 10^{13} 	ext{ electrons/s}$. The photoelectric current $I_p$ is given by $I_p = n_e 	imes e$,where $e = 1.6 	imes 10^{-19} \ C$: $I_p = 1.1475 	imes 10^{13} 	imes 1.6 	imes 10^{-19} \approx 1.836 	imes 10^{-6} \ A$. Rounding to the provided option,the correct value is $1.856 	imes 10^{-6} \ A$.

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