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: Plan:
Use kinetic energies and then take the square root.

Step 2: Steps:
The kinetic energies are in the ratio $1:4$, because $f - f_0$ is $f_0$ and $4f_0$. Speed varies as the square root of kinetic energy, so $v_1:v_2 = \sqrt1:\sqrt4 = 1:2$.

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