Q108 / 150 · MHT CET 2026 (11 April Shift 2) · Physics
Photoelectrons are emitted from two similar metal plates when wavelengths $\lambda_1$ and $\lambda_2$ are incident on them ($\lambda_1 = 1.5\lambda_2$). If maximum kinetic energy of emitted photoelectrons is $E_1$ and $E_2$ respectively then
- Correct answerA$E_1 < 2E_2/3$
- B$E_1 = 2E_2/3$
- C$E_1 = E_2/3$
- D$E_1 = 2E_2$
Solution
Using Einstein's photoelectric equation:
$E_1 = \dfrac{hc}{\lambda_1} - \phi$
$E_2 = \dfrac{hc}{\lambda_2} - \phi$
Given $\lambda_1 = 1.5\lambda_2 = \dfrac{3}{2}\lambda_2$, we can write:
$E_1 = \dfrac{2hc}{3\lambda_2} - \phi$
From the second equation, $\dfrac{hc}{\lambda_2} = E_2 + \phi$. Substituting this into the equation for $E_1$:
$E_1 = \dfrac{2}{3}(E_2 + \phi) - \phi$
$E_1 = \dfrac{2}{3}E_2 - \dfrac{1}{3}\phi$
Since the work function $\phi > 0$, it follows that:
$E_1 < \dfrac{2}{3}E_2$
Answer: $E_1 < 2E_2/3$
$E_1 = \dfrac{hc}{\lambda_1} - \phi$
$E_2 = \dfrac{hc}{\lambda_2} - \phi$
Given $\lambda_1 = 1.5\lambda_2 = \dfrac{3}{2}\lambda_2$, we can write:
$E_1 = \dfrac{2hc}{3\lambda_2} - \phi$
From the second equation, $\dfrac{hc}{\lambda_2} = E_2 + \phi$. Substituting this into the equation for $E_1$:
$E_1 = \dfrac{2}{3}(E_2 + \phi) - \phi$
$E_1 = \dfrac{2}{3}E_2 - \dfrac{1}{3}\phi$
Since the work function $\phi > 0$, it follows that:
$E_1 < \dfrac{2}{3}E_2$
Answer: $E_1 < 2E_2/3$
Hence, the correct answer is option: 'A'.