Photoelectric emission is observed from a metallic surface for frequencies \(ν_1\) and \(ν_2\) of the incident light rays (\(ν_1 > ν_2\)). If the ratio of maximum value of kinetic energy of the photoelectron emitted in first case to that in second case 3 : K, then the threshold frequency of the metallic surface is
Show Hint
- \(\frac{Kν_1-ν_2}{K-1}\)
- \(\frac{K-1}{Kν_1-ν_2}\)
- \(\frac{K-3}{Kν_1-3ν_2}\)
- \(\frac{Kν_1-3ν_2}{K-3}\)
The Correct Option is D
Solution and Explanation
Einstein's photoelectric equation: \(K_{max}=h\nu-h\nu_0=h(\nu-\nu_0)\).
Step 2: Key Formula or Approach
\[ \frac{K_1}{K_2}=\frac{\nu_1-\nu_0}{\nu_2-\nu_0}=\frac3K \]
Step 3: Detailed Explanation
Cross-multiply: \(K(\nu_1-\nu_0)=3(\nu_2-\nu_0)\).
\(K\nu_1-K\nu_0=3\nu_2-3\nu_0\), so \(K\nu_1-3\nu_2=(K-3)\nu_0\).
\[ \nu_0=\frac{K\nu_1-3\nu_2}{K-3} \]
Final Answer:
The threshold frequency is \(\frac{K\nu_1-3\nu_2}{K-3}\), option (D). \[ \boxed{\dfrac{K\nu_1-3\nu_2}{K-3}\ \text{(D)}} \]
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