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  • Calculate the kinetic energy of emitted photoelectrons, when a photon of light with falls on a meta

Calculate the kinetic energy of emitted photoelectrons, when a photon of light with \lambda=400 \mathrm{~nm} falls on a metal surface. with ejection of photoelectrons with a velocity of 4.4 \times 10^5 \mathrm{~m} / \mathrm{s}.

Option: 1

1.16 eV


Option: 2

2.8 eV


Option: 3

0.55 eV


Option: 4

9.1 eV


Answers (1)

best_answer

\begin{aligned} \mathrm{K \cdot E \text { of electron } }& =\frac{1}{2} \mathrm{mV}^2 \\ \\& =\frac{1}{2}\left(9.1 \times 10^{-31}\right)\left(4.4 \times 10^5\right) \\ \\& =0.55 \mathrm{eV} \end{aligned}

\Rightarrow Option (3) is the right answer

Posted by

Gaurav

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