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CUET · PHYSICS · PYQ PAPER 2023

The work function of cesium is 2.27 eV. The cut off voltage which stops the emission of electron from cesium when irradiated with light of wavelength 400 nm is

  1. A 0.84 V
  2. B 0.94 V
  3. C 1.2 V
  4. D 1.8 V
Verified Solution

Answer & Solution

Correct Answer

(A) 0.84 V

Step-by-step Solution

Detailed explanation

\(E = \frac{hc}{\lambda}\) \(E = \frac{1240 \text{ eV nm}}{400 \text{ nm}} = 3.1 \text{ eV}\) \(V_0 = E - \phi\) \(V_0 = 3.1 \text{ eV} - 2.27 \text{ eV} = 0.83 \text{ V}\)
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