Question

Two identical photocathodes receive light of frequencies \(n_1\) and \(n_2\). If the velocities of the emitted photoelectrons of mass m are \(V_1\) and \(V_2\) respectively, then (\(h\) = Planck's constant)

• A. \(V_1+V_2 = [\frac{2h}{m}(n_1+n_2)]^{\frac{1}{2}}\)
• B. \(V_1-V_2 = [\frac{2h}{m}(n_1-n_2)]^{\frac{1}{2}}\)
• C. \(V_1^2+V_2^2 = \frac{2h}{m}(n_1+n_2)\)
• D. \(V_1^2-V_2^2 = \frac{2h}{m}(n_1-n_2)\)

Hint:

Use Einstein's equation for the two identical cathodes and subtract.

Answer 1

The Correct Option is D

Step 1: Eliminate the work function:
The difference of the two kinetic energies is the difference of the photon energies, since $\phi$ cancels.

Step 2: Write it:
$\frac12mV_1^2 - \frac12mV_2^2 = hn_1 - hn_2$.

Step 3: Rearrange:
Multiply by $\frac2m$: $V_1^2 - V_2^2 = \frac{2h}{m}(n_1 - n_2)$.

Final Answer:
Option (D) is correct. \[ \boxed{V_1^2-V_2^2=\frac{2h}{m}(n_1-n_2)} \]