MHT CET · Physics · Dual Nature of Matter
Energy of the incident photon on the metal surface is ' \(3 \mathrm{~W}\) ' and then ' \(5 \mathrm{~W}\) ', where 'W' is the work function for that metal. The ratio of velocities of emitted photoelectrons is
- A \(1: \sqrt{2}\)
- B \(1: 1\)
- C \(1: 2\)
- D \(1: 4\)
Answer & Solution
Correct Answer
(A) \(1: \sqrt{2}\)
Step-by-step Solution
Detailed explanation
\( \begin{aligned} & \frac{1}{2} m v_{1}^{2}=3 W-W=2 W \\ & \frac{1}{2} m v_{2}^{2}=5 W-W=4 W \\ & \frac{v_{1}^{2}}{v_{2}^{2}}=\frac{1}{2} \\ \therefore & \frac{v_{1}}{v_{2}}=\frac{1}{\sqrt{2}} \end{aligned} \)
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