MHT CET · Physics · Dual Nature of Matter
The maximum velocity of the photoelectrons emitted by a metal surface is \(9 \times 10^5 \mathrm{~m} / \mathrm{s}\). The value of ratio of charge (e) to mass (m) of the photoelectron is \(1.8 \times 10^{11} \mathrm{C} / \mathrm{kg}\). The value of stopping potential in volt is
- A \(2 \cdot 00\)
- B \(2 \cdot 25\)
- C \(2 \cdot 50\)
- D \(3.00\)
Answer & Solution
Correct Answer
(B) \(2 \cdot 25\)
Step-by-step Solution
Detailed explanation
\( V_s = \frac{v^2}{2(e/m)} \) \( V_s = \frac{(9 \times 10^5)^2}{2 \times (1.8 \times 10^{11})} \)
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