Photoelectric emission is observed from a metallic surface for frequencies \(v_1\) and \(v_2\) of the incident light rays \(\left(v_1>v_2\right)\). If the maximum values of kinetic energy of the photoelectrons emitted in the two cases are in the ratio of \(1: \mathrm{k}\), then the threshold frequency of metallic surface is
Photoelectric emission is observed from a metallic surface for frequencies \(v_1\) and \(v_2\) of the incident light rays \(\left(v_1>v_2\right)\). If the maximum values of kinetic energy of the photoelectrons emitted in the two cases are in the ratio of \(1: \mathrm{k}\), then the threshold frequency of metallic surface is
Show Hint
- \(\frac{\mathrm{kv}_{2}-\mathrm{v}_{1}}{\mathrm{k}-1}\)
- \(\frac{v_{2}-v_{1}}{k}\)
- \(\frac{v_{1}-v_{2}}{k-1}\)
- \(\frac{\mathrm{kv}_{1}-\mathrm{v}_{2}}{\mathrm{k}-1}\)
The Correct Option is D
Solution and Explanation
Einstein's equation: \(KE=h(\nu-\nu_0)\).
Step 2: Ratio
\(\dfrac{KE_1}{KE_2}=\dfrac{\nu_1-\nu_0}{\nu_2-\nu_0}=\dfrac{1}{k}\)
Step 3: Solve
\(k(\nu_1-\nu_0)=\nu_2-\nu_0\Rightarrow k\nu_1-\nu_2=(k-1)\nu_0\)
\(\nu_0=\dfrac{k\nu_1-\nu_2}{k-1}\), option (D).
Final Answer:
The threshold frequency is (k nu1 minus nu2) over (k minus 1). \[ \boxed{\text{(D) }\dfrac{kv_{1}-v_{2}}{k-1}} \]
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