Question:
From a metallic surface photoelectric emission is observed for frequencies $v_{1}$ and $v_{2}$ ($v_{1}>v_{2}$) of the incident light. The maximum values of the kinetic energy of the photoelectrons emitted in the two cases are in the ratio $1:x$. Hence the threshold frequency of the metallic surface is
From a metallic surface photoelectric emission is observed for frequencies $v_{1}$ and $v_{2}$ ($v_{1}>v_{2}$) of the incident light. The maximum values of the kinetic energy of the photoelectrons emitted in the two cases are in the ratio $1:x$. Hence the threshold frequency of the metallic surface is
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Logic Tip: Threshold frequency is a characteristic property of the metal surface and does not depend on the incident light frequency.
Updated On: Apr 28, 2026
- $\frac{v_{1}-v_{2{x}$
- $\frac{v_{1}-v_{2{x-1}$
- $\frac{xv_{1}-v_{2{x-1}$
- $\frac{xv_{2}-v_{1{x-1}$
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The Correct Option is C
Solution and Explanation
Concept:
According to Einstein’s photoelectric equation:
\[
E_k = h(\nu - \nu_0)
\]
where $\nu_0$ is the threshold frequency.
Step 1: Write equations for both cases \[ E_{k1} = h(\nu_1 - \nu_0), \qquad E_{k2} = h(\nu_2 - \nu_0) \]
Step 2: Use the given ratio \[ \frac{E_{k1{E_{k2 = \frac{1}{x} \] Substituting: \[ \frac{\nu_1 - \nu_0}{\nu_2 - \nu_0} = \frac{1}{x} \]
Step 3: Solve for $\nu_0$ \[ x(\nu_1 - \nu_0) = \nu_2 - \nu_0 \] \[ x\nu_1 - x\nu_0 = \nu_2 - \nu_0 \] \[ x\nu_1 - \nu_2 = x\nu_0 - \nu_0 \] \[ x\nu_1 - \nu_2 = \nu_0 (x - 1) \] \[ \Rightarrow \nu_0 = \frac{x\nu_1 - \nu_2}{x - 1} \]
Step 4: Final Answer \[ \boxed{\nu_0 = \frac{x\nu_1 - \nu_2}{x - 1 \]
Step 1: Write equations for both cases \[ E_{k1} = h(\nu_1 - \nu_0), \qquad E_{k2} = h(\nu_2 - \nu_0) \]
Step 2: Use the given ratio \[ \frac{E_{k1{E_{k2 = \frac{1}{x} \] Substituting: \[ \frac{\nu_1 - \nu_0}{\nu_2 - \nu_0} = \frac{1}{x} \]
Step 3: Solve for $\nu_0$ \[ x(\nu_1 - \nu_0) = \nu_2 - \nu_0 \] \[ x\nu_1 - x\nu_0 = \nu_2 - \nu_0 \] \[ x\nu_1 - \nu_2 = x\nu_0 - \nu_0 \] \[ x\nu_1 - \nu_2 = \nu_0 (x - 1) \] \[ \Rightarrow \nu_0 = \frac{x\nu_1 - \nu_2}{x - 1} \]
Step 4: Final Answer \[ \boxed{\nu_0 = \frac{x\nu_1 - \nu_2}{x - 1 \]
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