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  • One milliwatt of light of wavelength  is incident on a caesium surface. What is the photoelectr

One milliwatt of light of wavelength 4560\, \AA is incident on a caesium surface. What is the photoelectric current produced, if the quantum efficiency of the surface for photoelectric emission is only 0.5%.

Option: 1

2.3 \times 10^{-6} \mathrm{~A}


Option: 2

0.84 \times 10^{-6} \mathrm{~A}


Option: 3

1.84 \times 10^{-6} \mathrm{~A}


Option: 4

None of these


Answers (1)

best_answer

Given that  \lambda=4560 \AA

=4560 \times 10^{-10} \mathrm{~m}

E= Energy of one photon of incident light of wavelength \lambda

=\frac{\mathrm{hc}}{\lambda}=\frac{6.6 \times 10^{-34} \times 3 \times 10^8}{4560 \times 10^{-10}}

=4.34 \times 10^{19} \mathrm{~J}

Because 1 \mathrm{~mW} of light of wavelength \lambda is incident on the caesium surface, hence number of photon falling per sec. on the surface

\begin{gathered} \mathrm{n}=\frac{\mathrm{P}}{(\mathrm{hc} / \lambda)}=\frac{10^{-3} \text { Watt }}{4.34 \times 10^{-19}} \\ =2.35 \times 10^{15} \end{gathered}

As quantum efficiency of the surface is only 0.5 \% hence only 0.5 \% of these photons liberate photoelectrons i.e. number of electrons liberated from the surface/sec.

n^{\prime}=n \times \frac{0.5}{100}=2.35 \times 10^{15} \times \frac{0.5}{100}=1.15 \times 10^{13}

Hence photoelectric current 

 \mathrm{i}=\mathrm{n}^{\prime} \mathrm{e}=1.15 \times 10^{13} \times 1.6 \times 10^{-19} \mathrm{amp}

=1.84 \times 10^{-6} \mathrm{amp}

Posted by

himanshu.meshram

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