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: Understanding the Concept:
Einstein's photoelectric equation is \(h\nu = \phi + \frac12mV^2\). Both cathodes are identical, so the work function \(\phi\) is the same.

Step 2: Write both equations:
\(hn_1 = \phi + \frac12mV_1^2\) and \(hn_2 = \phi + \frac12mV_2^2\).

Step 3: Subtract:
\[ h(n_1 - n_2) = \frac12m(V_1^2 - V_2^2) \Rightarrow V_1^2 - V_2^2 = \frac{2h}{m}(n_1 - n_2) \]

Step 4: Why the other options are wrong.
Options with \(V_1 + V_2\) or \(V_1 - V_2\) have the speeds themselves rather than their squares. Option (C) adds the equations, which would leave \(2\phi\) in the result.

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