MHT CET · Physics · Dual Nature of Matter
A photosensitive surface has work function \(\phi\). If photon of energy \(3 \phi\) falls on this surface, the electron comes out with maximum velocity of \(4 \times 10^6 \mathrm{~m} / \mathrm{s}\). When photon energy is increased to \(7 \phi\) then maximum velocity of photoelectron will be
- A \(4 \sqrt{3} \times 10^6 \mathrm{~m} / \mathrm{s}\)
- B \(2 \sqrt{3} \times 10^6 \mathrm{~m} / \mathrm{s}\)
- C \(4 \sqrt{3} \times 10^3 \mathrm{~m} / \mathrm{s}\)
- D \(2 \sqrt{3} \times 10^3 \mathrm{~m} / \mathrm{s}\)
Answer & Solution
Correct Answer
(A) \(4 \sqrt{3} \times 10^6 \mathrm{~m} / \mathrm{s}\)
Step-by-step Solution
Detailed explanation
\(KE_{max1} = E_1 - \phi = 3\phi - \phi = 2\phi\) \(\frac{1}{2} m v_1^2 = 2\phi \implies mv_1^2 = 4\phi\)
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