AP EAMCET · PHYSICS · Dual Nature of Matter
Photoelectric emission is observed from a metallic surface for frequencies \(v_1\) and \(v_2\) of the incident light \(\left(v_1>v_2\right)\). If the maximum values of kinetic energy of the photoelectrons emitted in the two cases are in the ratio \(1: n\), then the threshold frequency of the metallic surface is
- A \(\frac{\left(v_1-v_2\right)}{(n-1)}\)
- B \(\frac{\left(n v_1-v_2\right)}{(n-1)}\)
- C \(\frac{\left(n v_2-v_1\right)}{(n-1)}\)
- D \(\frac{\left(v_1-v_2\right)}{n}\)
Answer & Solution
Correct Answer
(B) \(\frac{\left(n v_1-v_2\right)}{(n-1)}\)
Step-by-step Solution
Detailed explanation
By using \(h v-h v_0=K_{\max }\) \(h\left(v_1-v_0\right)=K_1\) \(h\left(v_2-v_0\right)=K_2\) \(\therefore \quad \frac{v_1-v_0}{v_2-v_0}=\frac{K_1}{K_2}=\frac{1}{n}\) or \(\quad v_0=\frac{n v_1-v_2}{(n-1)}\)
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