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: Understanding the Question:
The question uses Einstein's photoelectric equation to relate incident photon energy and kinetic energy of emitted electrons.
Step 2: Key Formula or Approach:
Einstein's photoelectric equation: \(KE_{max} = \frac{1}{2}mv^2 = h\nu - \phi\).
Here \(\phi = W\). So, \( \frac{1}{2}mv^2 = E - W\).
Step 3: Detailed Explanation:
Case 1: Initial incident energy \(E_1 = 4W\).
\[ \frac{1}{2}mv_1^2 = 4W - W = 3W \quad \dots (i) \]
Case 2: Final incident energy \(E_2 = 6W\).
\[ \frac{1}{2}mv_2^2 = 6W - W = 5W \quad \dots (ii) \]
Divide equation (i) by (ii):
\[ \frac{v_1^2}{v_2^2} = \frac{3W}{5W} = \frac{3}{5} \]
Taking the square root on both sides:
\[ \frac{v_1}{v_2} = \frac{\sqrt{3}}{\sqrt{5}} \]
The ratio is \(\sqrt{3} : \sqrt{5}\).
Step 4: Final Answer:
The ratio of velocities is \(\sqrt{3} : \sqrt{5}\).