A light of wavelength λ(λ < λ_0) falls on a metal surface. The...

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

Options

A. $\dfrac{\lambda\lambda_0}{4\lambda_0 - 3\lambda}$

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)<br><br>Applying the photoelectric equation again:<br><br>4 \left( \dfrac{p^2}{2m} \right) = \dfrac{hc}{\lambda'} - \dfrac{hc}{\lambda_0}<br><br>Substituting the value of \dfrac{p^2}{2m} from the first equation:<br><br>4 \left( \dfrac{hc}{\lambda} - \dfrac{hc}{\lambda_0} \right) = \dfrac{hc}{\lambda'} - \dfrac{hc}{\lambda_0}<br><br>Dividing both sides by

:<br><br>\dfrac{4}{\lambda} - \dfrac{4}{\lambda_0} = \dfrac{1}{\lambda'} - \dfrac{1}{\lambda_0}<br><br>\dfrac{1}{\lambda'} = \dfrac{4}{\lambda} - \dfrac{3}{\lambda_0}<br><br>\dfrac{1}{\lambda'} = \dfrac{4\lambda_0 - 3\lambda}{\lambda\lambda_0}<br><br>

\lambda' =

\dfrac{\lambda\lambda_0}{4\lambda_0 - 3\lambda}<br><br>Answer: \dfrac{\lambda\lambda_0}{4\lambda_0 - 3\lambda}