Question

Two identical photocathodes receive light of frequencies ' \(f 1\) ' and ' \(f_2\) '. The velocity of the photoelectrons of mass ' \(m\) ' emitted are respectively ' \(\mathrm{v}_1\) ' and ' \(\mathrm{v}_2\) '. Then the correct relation is ( \(\mathrm{h}=\) Planck's constant)

• A. \(v_{1}-v_{2}=[\frac{2h}{m}(f_{1}-f_{2})]^{\frac{1}{2}}\)
• B. \(v_{1}+v_{2}=[\frac{2h}{m}(f_{1}+f_{2})]^{\frac{1}{2}}\)
• C. \(v_{1}^{2}-v_{2}^{2}=\frac{2h}{m}(f_{1}-f_{2})\)
• D. \(v_{1}^{2}+v_{2}^{2}=\frac{2h}{m}(f_{1}+f_{2})\)

Hint:

Write Einstein's photoelectric equation for each frequency and subtract to remove the work function.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
Einstein's equation: \(hf=\phi+\dfrac{1}{2}mv^{2}\). The photocathodes are identical, so the work function \(\phi\) is the same.

Step 2: Write for both frequencies
\(hf_1=\phi+\dfrac{1}{2}mv_1^{2}\)
\(hf_2=\phi+\dfrac{1}{2}mv_2^{2}\)

Step 3: Subtract
\(h(f_1-f_2)=\dfrac{1}{2}m\left(v_1^{2}-v_2^{2}\right)\)
Hence \(v_1^{2}-v_2^{2}=\dfrac{2h}{m}(f_1-f_2)\).

Step 4: Compare the options
This is option (C).

Final Answer:
The correct relation is \(v_1^{2}-v_2^{2}=\dfrac{2h}{m}(f_1-f_2)\). \[ \boxed{\text{(C) }v_{1}^{2}-v_{2}^{2}=\dfrac{2h}{m}(f_{1}-f_{2})} \]