Question

The threshold frequency of metal is \(f_0\). When the light of frequency \(2f_0\) is incident on the metal plate, the maximum velocity of photoelectron is \(v_1\). When the frequency of incident radiation is increased to \(5f_0\) the maximum velocity of photoelectrons emitted is \(v_2\). The ratio \(v_1\) to \(v_2\) is

• A. \(1:2\)
• B. \(1:8\)
• C. \(1:16\)
• D. \(1:4\)

Hint:

\(\frac12mv^2=h(f-f_0)\), so \(v\propto\sqrt{f-f_0}\).

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Einstein's equation: \(\frac12mv_{\max}^2 = h(f - f_0)\).

Step 2: Two cases:
At \(f = 2f_0\): \(\frac12mv_1^2 = h(2f_0 - f_0) = hf_0\).
At \(f = 5f_0\): \(\frac12mv_2^2 = h(5f_0 - f_0) = 4hf_0\).

Step 3: Ratio:
\[ \frac{v_1^2}{v_2^2} = \frac{1}{4} \Rightarrow \frac{v_1}{v_2} = \frac12 \]
So \(v_1:v_2 = 1:2\). A ratio of \(1:4\) would result from forgetting the square root.

Final Answer:
The ratio \(v_1:v_2\) is \(1:2\), option (A). \[ \boxed{1:2} \]