Question:

Work function of Cu metal is 4.8 eV. What is the approximate wavelength of incident radiation required to eject electrons from its surface? ($h = 6.626 \times 10^{-34}$ Js)

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To eject electrons, use $\lambda = \dfrac{hc}{\phi}$ where $\phi$ is in joules.
Updated On: May 12, 2025
  • 500 nm
  • 450 nm
  • 400 nm
  • 258 nm
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The Correct Option is D

Solution and Explanation

Work function $\phi = 4.8$ eV = $4.8 \times 1.6 \times 10^{-19}$ J = $7.68 \times 10^{-19}$ J
Using: $\phi = \dfrac{hc}{\lambda} \Rightarrow \lambda = \dfrac{hc}{\phi}$
Substitute: $h = 6.626 \times 10^{-34}$ Js, $c = 3 \times 10^8$ m/s
$\lambda = \dfrac{6.626 \times 10^{-34} \cdot 3 \times 10^8}{7.68 \times 10^{-19}} = 2.58 \times 10^{-7}$ m = 258 nm
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