Question:
The threshold frequency for a certain photosensitive metal surface is _0. When light of frequency 2_0 is incident on the surface, the maximum velocity of the emitted photoelectrons is v_1. If the frequency is increased to 5_0, the maximum velocity becomes v_2. The ratio v_1 : v_2 is:
The threshold frequency for a certain photosensitive metal surface is _0. When light of frequency 2_0 is incident on the surface, the maximum velocity of the emitted photoelectrons is v_1. If the frequency is increased to 5_0, the maximum velocity becomes v_2. The ratio v_1 : v_2 is:
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Photoelectric speed depends on excess frequency above threshold:
\[
v^2 \propto (\nu - \nu_0)
\]
Updated On: Jun 10, 2026
- \(1:2\)
- \(1:4\)
- \(2:1\)
- \(1:\sqrt{2}\)
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The Correct Option is A
Solution and Explanation
Concept:
According to Einstein’s photoelectric equation, the maximum kinetic energy of emitted photoelectrons is:
\[
K_{\max} = h\nu - h\nu_0
\]
and
\[
K_{\max} = \frac{1}{2}mv^2
\]
Step 1: First case ( = 2_0) \[ K_1 = h(2\nu_0 - \nu_0) = h\nu_0 \] \[ \frac{1}{2}mv_1^2 = h\nu_0 \]
Step 2: Second case ( = 5_0) \[ K_2 = h(5\nu_0 - \nu_0) = 4h\nu_0 \] \[ \frac{1}{2}mv_2^2 = 4h\nu_0 \]
Step 3: Ratio \[ \frac{v_1^2}{v_2^2} = \frac{1}{4} \Rightarrow \frac{v_1}{v_2} = \frac{1}{2} \] Thus, \[ v_1 : v_2 = 1 : 2 \]
Step 1: First case ( = 2_0) \[ K_1 = h(2\nu_0 - \nu_0) = h\nu_0 \] \[ \frac{1}{2}mv_1^2 = h\nu_0 \]
Step 2: Second case ( = 5_0) \[ K_2 = h(5\nu_0 - \nu_0) = 4h\nu_0 \] \[ \frac{1}{2}mv_2^2 = 4h\nu_0 \]
Step 3: Ratio \[ \frac{v_1^2}{v_2^2} = \frac{1}{4} \Rightarrow \frac{v_1}{v_2} = \frac{1}{2} \] Thus, \[ v_1 : v_2 = 1 : 2 \]
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