CUET · PHYSICS · PYQ PAPER 2023
When the light of frequency \(2 \nu_0\) (where \(\nu_0\) is the threshold frequency) is incident on a photosensitive material, the maximum velocity of electrons emitted is \(v_1\). When the frequency of the incident radiation is increased to \(3 \nu_0\), the maximum velocity of emitted electrons from the same material is \(v_2\). The ratio of \(v_1\) to \(v_2\) will then be :
- A \(2: 1\)
- B \(\sqrt{2}: 1\)
- C \(1: 2\)
- D \(1: \sqrt{2}\)
Answer & Solution
Correct Answer
(D) \(1: \sqrt{2}\)
Step-by-step Solution
Detailed explanation
\(\frac{1}{2}mv_1^2 = h(2\nu_0) - h\nu_0 = h\nu_0\) \(\frac{1}{2}mv_2^2 = h(3\nu_0) - h\nu_0 = 2h\nu_0\) \(\frac{v_1^2}{v_2^2} = \frac{h\nu_0}{2h\nu_0} = \frac{1}{2}\) \(\frac{v_1}{v_2} = \frac{1}{\sqrt{2}}\)
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