TS EAMCET · Physics · Dual Nature of Matter
Let \(v_1\) and \(v_2\) be the maximum velocities of the emitted electrons when the surface of a metal is illuminated with light waves of energy \(E_1=4 \mathrm{eV}\) and \(E_2=2.5 \mathrm{eV}\), respectively. If the work function of the metal is \(2 \mathrm{eV}\), then the ratio \(\frac{v_1}{v_2}\) is
- A 1.6
- B 4
- C 2
- D 0.5
Answer & Solution
Correct Answer
(C) 2
Step-by-step Solution
Detailed explanation
In case of photoemission, \(K_{\max }=\frac{1}{2} m v_{\max }^2=h f-\phi_0\) For photons of energies \(4 \mathrm{eV}\) and \(2.5 \mathrm{eV}\), we have ( given, work function \(=2 \mathrm{eV}\) ) \(\frac{1}{2} m v_1^2=4-2=2\) \(\ldots(i)\) and…
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