Question:
When a certain photosensitive surface is illuminated with monochromatic light of frequency v, the stopping potential for the current is \(-V_0/2\). When the surface is illuminated by monochromatic light of frequency v/2, the stopping potential is \(-V_0\). The threshold frequency for photoelectric emission is :
When a certain photosensitive surface is illuminated with monochromatic light of frequency v, the stopping potential for the current is \(-V_0/2\). When the surface is illuminated by monochromatic light of frequency v/2, the stopping potential is \(-V_0\). The threshold frequency for photoelectric emission is :
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Einstein: \(K_{max}=hv-hv_0=e|V_s|\). Take the two cases as printed and subtract (or use \(K_2=2K_1\)) to eliminate \(V_0\).
Updated On: Nov 5, 2026
- \(2v\)
- \(\frac{4}{3}v\)
- \(\frac{3v}{2}\)
- \(\frac{5v}{3}\)
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The Correct Option is C
Solution and Explanation
Let \(\phi=hv_0\). Taking the data as printed: \(hv-\phi=\frac{eV_0}{2}\) ... (1) and \(\frac{hv}{2}-\phi=eV_0\) ... (2).
Multiply (1) by 2: \(2hv-2\phi=eV_0\). Equate with (2): \(2hv-2\phi=\frac{hv}{2}-\phi\Rightarrow\phi=\frac{3hv}{2}\).
So \(hv_0=\frac{3hv}{2}\Rightarrow v_0=\frac{3v}{2}\).
Answer: (3) \(\frac{3v}{2}\)
Multiply (1) by 2: \(2hv-2\phi=eV_0\). Equate with (2): \(2hv-2\phi=\frac{hv}{2}-\phi\Rightarrow\phi=\frac{3hv}{2}\).
So \(hv_0=\frac{3hv}{2}\Rightarrow v_0=\frac{3v}{2}\).
Answer: (3) \(\frac{3v}{2}\)
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