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CUET · PHYSICS · PYQ PAPER 2023

When the surface of a metal is illuminated with the light of wavelength 400 nm, the KE of the ejected photoelectrons is 1.68 eV. Then the work function of the metal is (hc = 1240 eV-nm) :

  1. A \(3.1 eV\)
  2. B \(1.42 eV\)
  3. C \(1.51 eV\)
  4. D \(1.68 eV\)
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Answer & Solution

Correct Answer

(B) \(1.42 eV\)

Step-by-step Solution

Detailed explanation

\(\phi = \frac{hc}{\lambda} - KE\) \(\phi = \frac{1240 \text{ eV-nm}}{400 \text{ nm}} - 1.68 \text{ eV}\) \(\phi = 3.1 \text{ eV} - 1.68 \text{ eV}\) \(\phi = 1.42 \text{ eV}\)
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