COMEDK · Physics · 27. Dual Nature of Matter
Threshold wavelength for photoelectric emission from a metal surface is \(5200\) \(Å\). Photoelectrons will be emitted, when this surface is illuminated with monochromatic radiation from
- A \(1 \mathrm{~W~IR}\) lamp
- B \(50 \mathrm{~W~UV}\) lamp
- C \(50 \mathrm{~W~IR}\) lamp
- D \(10 \mathrm{~W~IR}\) lamp
Answer & Solution
Correct Answer
(B) \(50 \mathrm{~W~UV}\) lamp
Step-by-step Solution
Detailed explanation
Given, \(\lambda_{0}=5200 Å=5200 \times 10^{-10} \mathrm{~m}\) \(\therefore\) Threshold frequency, \(f_{0}=\frac{c}{\lambda_{0}}=\frac{3 \times 10^{8}}{5200 \times 10^{-10}}\) \(=5.76 \times 10^{14} \mathrm{~Hz}\) It lies in the frequency range of UV-rays, so…
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