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 _______.
- A Green
- B Blue
- C Red
- D Yellow
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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