Question

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

• A. \(\frac{Kν_1-ν_2}{K-1}\)
• B. \(\frac{K-1}{Kν_1-ν_2}\)
• C. \(\frac{K-3}{Kν_1-3ν_2}\)
• D. \(\frac{Kν_1-3ν_2}{K-3}\)

Hint:

Use \(K_{max}=h(\nu-\nu_0)\) for both frequencies.

Answer 1

The Correct Option is D

Step 1: Set up the ratio
$\dfrac{\nu_1-\nu_0}{\nu_2-\nu_0}=\dfrac3K$.

Step 2: Isolate $\nu_0$
$K\nu_1-3\nu_2=\nu_0(K-3)$, so $\nu_0=\dfrac{K\nu_1-3\nu_2}{K-3}$, option (D).

Final Answer:
Option (D) is the threshold frequency. \[ \boxed{\dfrac{K\nu_1-3\nu_2}{K-3}} \]