Q186 / 200 · MHT CET - Medical 2026 (25 April Shift 2) · Physics

A light of wavelength $\lambda$ $(\lambda < \lambda_0)$ falls on a metal surface. The momentum of the fastest moving electron is $p$. In order to have the momentum of an electron equal to $2p$, the required wavelength $\lambda'$ of incident light must be

  • Correct answer
    A
    $\dfrac{\lambda\lambda_0}{4\lambda_0 - 3\lambda}$
  • B
    $\dfrac{4\lambda_0 - 3\lambda}{\lambda\lambda_0}$
  • C
    $\dfrac{\lambda_0}{4\lambda_0 - 3\lambda}$
  • D
    $\dfrac{4\lambda_0 - 3\lambda}{\lambda}$

Solution

Using Einstein's photoelectric equation:

$K_{\max} = \dfrac{hc}{\lambda} - \dfrac{hc}{\lambda_0}$

Since $K_{\max} = \dfrac{p^2}{2m}$, we have:

$\dfrac{p^2}{2m} = \dfrac{hc}{\lambda} - \dfrac{hc}{\lambda_0}$

For the new wavelength $\lambda'$, the momentum is $2p$. The new maximum kinetic energy is:

$K'_{\max} = \dfrac{(2p)^2}{2m} = 4 \left( \dfrac{p^2}{2m} \right)$

Applying the photoelectric equation again:

$4 \left( \dfrac{p^2}{2m} \right) = \dfrac{hc}{\lambda'} - \dfrac{hc}{\lambda_0}$

Substituting the value of $\dfrac{p^2}{2m}$ from the first equation:

$4 \left( \dfrac{hc}{\lambda} - \dfrac{hc}{\lambda_0} \right) = \dfrac{hc}{\lambda'} - \dfrac{hc}{\lambda_0}$

Dividing both sides by $hc$:

$\dfrac{4}{\lambda} - \dfrac{4}{\lambda_0} = \dfrac{1}{\lambda'} - \dfrac{1}{\lambda_0}$

$\dfrac{1}{\lambda'} = \dfrac{4}{\lambda} - \dfrac{3}{\lambda_0}$

$\dfrac{1}{\lambda'} = \dfrac{4\lambda_0 - 3\lambda}{\lambda\lambda_0}$

$\lambda' = \dfrac{\lambda\lambda_0}{4\lambda_0 - 3\lambda}$

Answer: $\dfrac{\lambda\lambda_0}{4\lambda_0 - 3\lambda}$

Hence, the correct answer is option: 'A'.

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