Question

If the momentum of an electron is changed by $\Delta p$ , then the de-Broglie wavelength associated with it changes by $0.05\%$ . The initial momentum of the electron will be:
(A) $\dfrac{{\Delta p}}{{200}}$
(B) $\dfrac{{\Delta p}}{{199}}$
(C) $199\Delta p$
(D) $- 2000\Delta p$

Answer 1

As the wavelength is inversely proportional to the momentum, so we can set the ratio of the initial momentum to the final momentum in terms of the ratio of the final wavelength to the initial wavelength. We can set the final momentum as the initial momentum plus $\Delta p$ and we can set the final wavelength as the initial plus $0.05\%$ of the initial wavelength.
Formula Used: In this solution we will be using the following formula,
$\Rightarrow \lambda = \dfrac{h}{p}$
where $\lambda$ is the de-Broglie wavelength,
$h$ is the Planck’s constant and
$p$ is the momentum.

Complete step by step answer:
The de-Broglie wavelength of any electron is given by the formula,
$\Rightarrow \lambda = \dfrac{h}{p}$
Now since the Planck’s constant doesn’t change so we can write the wavelength is inversely proportional to the momentum of the electron.
Therefore we get,
$\Rightarrow \lambda \propto \dfrac{1}{p}$
Now let us consider the initial wavelength as ${\lambda _1}$ and the final wavelength as ${\lambda _2}$ . And let us consider the initial momentum to be, ${p_1}$ and the final momentum be ${p_2}$ . So we can write the equations for the initial and final conditions as,
$\Rightarrow {\lambda _1} \propto \dfrac{1}{{{p_1}}}$ and
$\Rightarrow {\lambda _2} \propto \dfrac{1}{{{p_2}}}$
So taking the ratio of the wavelengths,
$\Rightarrow \dfrac{{{\lambda _1}}}{{{\lambda _2}}} = \dfrac{{{p_2}}}{{{p_1}}}$ …….(1)
It is given in the question that the final wavelength changes by $0.05\%$ , so we can write, $\Rightarrow {\lambda _2} = {\lambda _1} + \dfrac{{0.05}}{{100}}{\lambda _1}$
Taking LCM as 100,
$\Rightarrow {\lambda _2} = \dfrac{{100 + 0.05}}{{100}}{\lambda _1}$
Hence we get,
$\Rightarrow {\lambda _2} = \dfrac{{100.05}}{{100}}{\lambda _1}$
And it is said that the final momentum increases by $\Delta p$ . So we can write,
$\Rightarrow {p_2} = {p_1} + \Delta p$
Now substituting these values in the equation (1) we get,
$\Rightarrow \dfrac{{100{\lambda _1}}}{{100.05{\lambda _1}}} = \dfrac{{{p_1} + \Delta p}}{{{p_1}}}$
In the LHS we can cancel the ${\lambda _1}$ from both the numerator and the denominator and in the RHS, we divide the fraction as,
$\Rightarrow \dfrac{{100}}{{100.05}} = 1 + \dfrac{{\Delta p}}{{{p_1}}}$
Now bringing the 1 to the LHS,
$\Rightarrow \dfrac{{100}}{{100.05}} - 1 = \dfrac{{\Delta p}}{{{p_1}}}$
Taking $100.05$ as LCM in the LHS we have,
$\Rightarrow \dfrac{{100 - 100.05}}{{100.05}} = \dfrac{{\Delta p}}{{{p_1}}}$
We can now take only ${p_1}$ to the LHS and bring everything else to the RHS an get,
$\Rightarrow {p_1} = \dfrac{{100.05}}{{ - 0.05}}\Delta p$
On calculating this we get a value of,
$\Rightarrow {p_1} = - 2001\Delta p$
So we can write this as approximately,
$\Rightarrow {p_1} = - 2000\Delta p$
Hence, option (D) is correct.

Note:
The de-Broglie wavelength is the probability density of finding an object at any given point in configuration space. According to the wave-particle duality, the de-Broglie wavelength is manifested in all objects according to quantum mechanics.