Question

A photosensitive surface has work function (\phi). If photon of energy (3\phi) falls on this surface, the electron comes out with maximum velocity of (4 \times 10^6 m/s). When photon energy is increased to (7\phi) then maximum velocity will be

• A. (4\sqrt{3} \times 10^6 m/s)
• B. (2\sqrt{3} \times 10^6 m/s)
• C. (4\sqrt{3} \times 10^3 m/s)
• D. (2\sqrt{3} \times 10^3 m/s)

Hint:

Maximum kinetic energy is directly proportional to (Photon energy - Work function).

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The question is based on Einstein's photoelectric equation. We need to find the new maximum velocity of photoelectrons when the incident energy is changed.
Step 2: Key Formula or Approach:
Einstein's photoelectric equation: \(K_{\text{max}} = \frac{1}{2} m v_{\text{max}}^2 = E - \phi\)
Where \(E\) is the photon energy and \(\phi\) is the work function.
Step 3: Detailed Explanation:
Case 1: Incident energy \(E_1 = 3\phi\).
\[ \frac{1}{2} m v_1^2 = 3\phi - \phi = 2\phi \quad \dots \text{(i)} \]
Given \(v_1 = 4 \times 10^6 \text{ m/s}\).
Case 2: Incident energy \(E_2 = 7\phi\).
\[ \frac{1}{2} m v_2^2 = 7\phi - \phi = 6\phi \quad \dots \text{(ii)} \]
Dividing (ii) by (i):
\[ \frac{v_2^2}{v_1^2} = \frac{6\phi}{2\phi} = 3 \]
Taking the square root:
\[ \frac{v_2}{v_1} = \sqrt{3} \implies v_2 = \sqrt{3} v_1 \]
Substitute \(v_1 = 4 \times 10^6 \text{ m/s}\):
\[ v_2 = \sqrt{3} \times (4 \times 10^6) = 4\sqrt{3} \times 10^6 \text{ m/s} \]
Step 4: Final Answer:
The maximum velocity will be \(4\sqrt{3} \times 10^6 \text{ m/s}\).