Question

Energy of the incident photons on the metal surface is initially \(4 \text{ W}\) and then \(6 \text{ W}\) where \(W\) is the work function of that metal. The ratio of velocities of emitted photoelectrons is

• A. \(\sqrt{3} : \sqrt{5}\)
• B. \(1 : 2\)
• C. \(2 : 3\)
• D. \(\sqrt{2} : \sqrt{3}\)

Hint:

Velocity ratio is the square root of the ratio of the excess energies ($E-W$).

Answer 1

The Correct Option is A

Step 1: Photoelectric Equation
$K.E._{max} = \frac{1}{2}mv^2 = E - W$.

Step 2: Case 1
$\frac{1}{2}mv_1^2 = 4W - W = 3W$.

Step 3: Case 2
$\frac{1}{2}mv_2^2 = 6W - W = 5W$.

Step 4: Calculation
$\frac{v_1^2}{v_2^2} = \frac{3W}{5W} = \frac{3}{5} \implies \frac{v_1}{v_2} = \sqrt{\frac{3}{5}}$.
Final Answer: (A)