Photoelectric emission is observed from a metallic surface for frequencies \(v_1\) and \(v_2\) of the incident light rays (\(v_1 > v_2\)). If the maximum values of kinetic energy of the photoelectrons emitted in the two cases are in the ratio of \(k:1\), then what is the threshold frequency of the metallic surface?
Show Hint
- \(\frac{v_1-v_2}{k}\)
- \(\frac{v_1-v_2}{k-1}\)
- \(\frac{kv_1-v_2}{k-1}\)
- \(\frac{kv_2-v_1}{k-1}\)
The Correct Option is D
Solution and Explanation
\(K_1=h(\nu_1-\nu_0)\) and \(K_2=h(\nu_2-\nu_0)\).
Step 2: Use the ratio
\(\frac{K_1}{K_2}=k\) gives \(\nu_1-\nu_0=k(\nu_2-\nu_0)\).
Step 3: Solve
\(\nu_1-\nu_0=k\nu_2-k\nu_0\), so \((k-1)\nu_0=k\nu_2-\nu_1\) and \(\nu_0=\frac{k\nu_2-\nu_1}{k-1}\). Option (D).
Final Answer:
The threshold frequency is \(\frac{k\nu_2-\nu_1}{k-1}\), option (D). \[ \boxed{\text{(D)}} \]
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