Question:
Two photons of wavelength \( \lambda \) and \(2\lambda\) are incident on a metal surface and emit photoelectrons of maximum kinetic energies \(3k\) and \(k\). Find work function of the metal.
Two photons of wavelength \( \lambda \) and \(2\lambda\) are incident on a metal surface and emit photoelectrons of maximum kinetic energies \(3k\) and \(k\). Find work function of the metal.
Updated On: Apr 13, 2026
- \( \dfrac{hc}{4\lambda} \)
- \( \dfrac{hc}{2\lambda} \)
- \( \dfrac{2hc}{3\lambda} \)
- \( \dfrac{hc}{\lambda} \)
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The Correct Option is A
Solution and Explanation
Concept:
According to the photoelectric equation,
\[
\frac{hc}{\lambda} = W + K_{\max}
\]
where
\[
W = \text{work function}
\]
\[
K_{\max} = \text{maximum kinetic energy}
\]
Step 1:Apply equation for wavelength \( \lambda \). \[ \frac{hc}{\lambda} = W + 3k \]
Step 2:Apply equation for wavelength \(2\lambda\). \[ \frac{hc}{2\lambda} = W + k \]
Step 3:Solve the equations. Subtract the second equation from the first: \[ \frac{hc}{\lambda} - \frac{hc}{2\lambda} = 2k \] \[ \frac{hc}{2\lambda} = 2k \] \[ k = \frac{hc}{4\lambda} \] Substitute in \[ \frac{hc}{2\lambda} = W + k \] \[ \frac{hc}{2\lambda} = W + \frac{hc}{4\lambda} \] \[ W = \frac{hc}{4\lambda} \] \[ \boxed{W = \frac{hc}{4\lambda}} \]
Step 1:Apply equation for wavelength \( \lambda \). \[ \frac{hc}{\lambda} = W + 3k \]
Step 2:Apply equation for wavelength \(2\lambda\). \[ \frac{hc}{2\lambda} = W + k \]
Step 3:Solve the equations. Subtract the second equation from the first: \[ \frac{hc}{\lambda} - \frac{hc}{2\lambda} = 2k \] \[ \frac{hc}{2\lambda} = 2k \] \[ k = \frac{hc}{4\lambda} \] Substitute in \[ \frac{hc}{2\lambda} = W + k \] \[ \frac{hc}{2\lambda} = W + \frac{hc}{4\lambda} \] \[ W = \frac{hc}{4\lambda} \] \[ \boxed{W = \frac{hc}{4\lambda}} \]
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