Question:
Radiations of two photons having energy three and nine times the work function of a metal are incident on the metal surface successively. The ratio of maximum velocities of photoelectrons emitted in two cases are:
Radiations of two photons having energy three and nine times the work function of a metal are incident on the metal surface successively. The ratio of maximum velocities of photoelectrons emitted in two cases are:
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In photoelectric effect, velocity depends on square root of kinetic energy, not directly on energy.
Updated On: Jun 20, 2026
- \(1:3\)
- \(1:\sqrt{3}\)
- \(1:4\)
- \(1:2\)
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The Correct Option is D
Solution and Explanation
Step 1: Use Einstein’s photoelectric equation.
Maximum kinetic energy of emitted electron: \[ K_{\max} = E - \phi \]
Step 2: First case (energy = 3φ).
\[ K_1 = 3\phi - \phi = 2\phi \] So, \[ \frac{1}{2}mv_1^2 = 2\phi \Rightarrow v_1^2 \propto 2\phi \]
Step 3: Second case (energy = 9φ).
\[ K_2 = 9\phi - \phi = 8\phi \] So, \[ \frac{1}{2}mv_2^2 = 8\phi \Rightarrow v_2^2 \propto 8\phi \]
Step 4: Take ratio of velocities.
\[ \frac{v_1}{v_2} = \sqrt{\frac{2\phi}{8\phi}} = \sqrt{\frac{1}{4}} = \frac{1}{2} \]
Step 5: Final conclusion.
\[ \boxed{1:2} \]
Maximum kinetic energy of emitted electron: \[ K_{\max} = E - \phi \]
Step 2: First case (energy = 3φ).
\[ K_1 = 3\phi - \phi = 2\phi \] So, \[ \frac{1}{2}mv_1^2 = 2\phi \Rightarrow v_1^2 \propto 2\phi \]
Step 3: Second case (energy = 9φ).
\[ K_2 = 9\phi - \phi = 8\phi \] So, \[ \frac{1}{2}mv_2^2 = 8\phi \Rightarrow v_2^2 \propto 8\phi \]
Step 4: Take ratio of velocities.
\[ \frac{v_1}{v_2} = \sqrt{\frac{2\phi}{8\phi}} = \sqrt{\frac{1}{4}} = \frac{1}{2} \]
Step 5: Final conclusion.
\[ \boxed{1:2} \]
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