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JEE Mains · Physics · STD 12 - 11. Dual nature of radiation and matter

The work function of a metal is 3 eV. The color of the visible light that is required to cause emission of photoelectrons is _______.

  1. A Green
  2. B Blue
  3. C Red
  4. D Yellow
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Answer & Solution

Correct Answer

(B) Blue

Step-by-step Solution

Detailed explanation

\(\begin{aligned} & (\mathrm{KE})_{\max }=\frac{\mathrm{hc}}{\lambda}-\phi \\ & \frac{\mathrm{hc}}{\lambda} \gt \phi[\text { for emission }] \\ & \lambda \lt \frac{\mathrm{hc}}{\phi} \Rightarrow \lambda \lt \frac{1242}{3} \mathrm{~nm} \end{aligned}\) So blue light option (B)
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