Question:
Two identical photo cathodes receive light of frequencies \( f_1 \) and \( f_2 \). If the velocity of photo electrons (of mass m) coming out are respectively \( V_1 \) & \( V_2 \) then:
Two identical photo cathodes receive light of frequencies \( f_1 \) and \( f_2 \). If the velocity of photo electrons (of mass m) coming out are respectively \( V_1 \) & \( V_2 \) then:
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Always look to eliminate constants like the work function by comparing two different states of the same physical system.
Updated On: Jun 9, 2026
- \( V_1^2 - V_2^2 = \frac{2h}{m}(f_1 - f_2) \)
- \( V_1 + V_2 = [\frac{2h}{m}(f_1 + f_2)]^{1/2} \)
- \( V_1^2 + V_2^2 = \frac{2h}{m}(f_1 + f_2) \)
- \( V_1 - V_2 = [\frac{2h}{m}(f_1 - f_2)]^{1/2} \)
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The Correct Option is A
Solution and Explanation
Concept:
Einstein's photoelectric equation relates the energy of a photon, the work function (\(\Phi\)) of the metal surface, and the maximum kinetic energy of the emitted photoelectrons. The equation is \( hf = \Phi + K.E._{max} \). The kinetic energy of an electron with mass \( m \) and velocity \( V \) is given by \( K.E. = \frac{1}{2} m V^2 \).
Step 1: Apply the photoelectric equation to both cases.
For the first light frequency \( f_1 \) and velocity \( V_1 \): $$ hf_1 = \Phi + \frac{1}{2} m V_1^2 \quad \text{--- (Equation 1)} $$ For the second light frequency \( f_2 \) and velocity \( V_2 \): $$ hf_2 = \Phi + \frac{1}{2} m V_2^2 \quad \text{--- (Equation 2)} $$
Step 2: Eliminate the work function (\(\Phi\)).
Subtract Equation 2 from Equation 1 to eliminate the constant work function: $$ hf_1 - hf_2 = (\Phi + \frac{1}{2} m V_1^2) - (\Phi + \frac{1}{2} m V_2^2) $$ $$ h(f_1 - f_2) = \frac{1}{2} m (V_1^2 - V_2^2) $$
Step 3: Isolate the velocity term.
Multiply both sides by 2 and divide by the mass \( m \): $$ V_1^2 - V_2^2 = \frac{2h}{m} (f_1 - f_2) $$ $$\boxed{V_1^2 - V_2^2 = \frac{2h}{m}(f_1 - f_2)}$$
Step 1: Apply the photoelectric equation to both cases.
For the first light frequency \( f_1 \) and velocity \( V_1 \): $$ hf_1 = \Phi + \frac{1}{2} m V_1^2 \quad \text{--- (Equation 1)} $$ For the second light frequency \( f_2 \) and velocity \( V_2 \): $$ hf_2 = \Phi + \frac{1}{2} m V_2^2 \quad \text{--- (Equation 2)} $$
Step 2: Eliminate the work function (\(\Phi\)).
Subtract Equation 2 from Equation 1 to eliminate the constant work function: $$ hf_1 - hf_2 = (\Phi + \frac{1}{2} m V_1^2) - (\Phi + \frac{1}{2} m V_2^2) $$ $$ h(f_1 - f_2) = \frac{1}{2} m (V_1^2 - V_2^2) $$
Step 3: Isolate the velocity term.
Multiply both sides by 2 and divide by the mass \( m \): $$ V_1^2 - V_2^2 = \frac{2h}{m} (f_1 - f_2) $$ $$\boxed{V_1^2 - V_2^2 = \frac{2h}{m}(f_1 - f_2)}$$
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