When two monochromatic light of frequency ν and \(\frac{ν}{2}\) are incident on a photoelectric metal, their stopping potential becomes \(\frac{V_s}{2}\) and Vs respectively. The threshold frequency for this metal is:
- 2 v
- 3 v
\(\frac{2}{3v}\)
\(\frac{3}{2v}\)
The Correct Option is D
Solution and Explanation
\(hv = W + 2v_0 e\)
\(2hv = W + \frac{v_0}{e}\)
on solving we get,\(W = \frac{3}{2}hv\)
\(h_0 = \frac{3}{2}hv\)
\(v_0 = \frac{3}{2}v\)
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Heisenberg’s Uncertainty Principle states that both the momentum and position of a particle cannot be determined simultaneously.