If the momentum of an electron is changed by P, then the de Broglie wavelength associated with it changes by $0.5 \%. $ The initial momentum of electron will be

Question

The variation of the stopping potential $ V_0 $ with the frequency $ \nu $ of the incident radiation for four metals A, B, C, and D is shown in the figure. For the same frequency of incident radiation producing photo-electrons in all metals, the kinetic energy of photo-electrons will be maximum for which metal?
variation of the stopping potential

Answer 1

To find the initial momentum of the electron when its de Broglie wavelength changes by 0.5% due to a change in momentum by $ P $, let's start by recalling the de Broglie wavelength formula:

$\lambda = \frac{h}{p}$

where:

$\lambda$ is the de Broglie wavelength,
$h$ is Planck's constant,
$p$ is the momentum of the particle.

According to the problem, the de Broglie wavelength changes by 0.5%, or:

$\frac{\Delta \lambda}{\lambda} = 0.5\%$

Since $\lambda = \frac{h}{p}$, the relation between changes in wavelength and momentum can be derived using differentiation. Differentiating $\lambda$ with respect to $p$, we get:

$\Delta \lambda = -\frac{h}{p^2} \Delta p$

In percentage terms for small changes:

$\frac{\Delta \lambda}{\lambda} = -\frac{\Delta p}{p}$

Given that:

$\frac{\Delta \lambda}{\lambda} = 0.5\% = 0.005$

Thus, we have:

$0.005 = -\frac{P}{p}$

Solving for $p$, the initial momentum:

$p = \frac{P}{0.005} = 200P$

Therefore, the initial momentum of the electron is 200P.

This confirms that the correct answer is:

200P

Answer 2