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
- A 0.84 V
- B 0.94 V
- C 1.2 V
- D 1.8 V
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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