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 \(4 \times 10^6 \mathrm{~m} / \mathrm{s}\) When photon energy is increased to \(7 \phi\) then maximum velocity of photoelectron will be

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

Hint:

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

Answer 1

The Correct Option is A

Step 1: Einstein's Equation
$K.E._{max} = \frac{1}{2}mv^2 = E - \phi$.

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

Step 3: Case 2
$\frac{1}{2}mv_2^2 = 7\phi - \phi = 6\phi$.

Step 4: Ratio
$\frac{v_2^2}{v_1^2} = \frac{6\phi}{2\phi} = 3 \implies \frac{v_2}{v_1} = \sqrt{3}$.
$v_2 = v_1\sqrt{3} = 4\sqrt{3} \times 10^6 \text{ m/s}$.
Final Answer: (A)