Question

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 the electron will be:
(A) 400P
(B) $\dfrac{P}{{200}}$
(C) 100P
(D) 200P

Answer 1

When changes are involved, the derivative of the quantities should be inspected. In the de Broglie equation, an increase in wavelength will due to a decrease in momentum.
Formula used: In this solution we will be using the following formulae;
$\lambda = \dfrac{h}{p}$ where $\lambda$ is the wavelength of a matter wave, $h$ is the Planck’s constant, and $p$ is the momentum of the particle.

Complete Step-by-Step Solution:
The momentum of an electron is said to change by an amount $P$, the corresponding change in wavelength is $0.5\%$. We are to find the initial momentum of the electron.
Now, we recall that the equation of de Broglie can be given as
$\lambda = \dfrac{h}{p}$ where $\lambda$ is the wavelength of a matter wave, $h$ is the Planck’s constant, and $p$ is the momentum of the particle.
Hence, for the initial wavelength ${\lambda _1}$, we let the corresponding momentum be ${p_1}$. Hence, the de Broglie equation for this state can be given as
${\lambda _1} = \dfrac{h}{{{p_1}}}$
For instantaneous change in the wavelength, we differentiate the de Broglie equation, as in
$\dfrac{{d\lambda }}{{dp}} = - \dfrac{h}{{{p^2}}}$
$\Rightarrow d\lambda = - \dfrac{h}{{{p^2}}}dp$
But $\lambda = \dfrac{h}{p}$
Hence, $d\lambda = - \dfrac{\lambda }{p}dp$
Hence, $\dfrac{{\Delta \lambda }}{\lambda } = \dfrac{{\Delta p}}{p}$
Then by replacement with initial wavelength and momentum, we have
$\dfrac{{\Delta \lambda }}{{{\lambda _1}}} = \dfrac{{\Delta p}}{{{p_1}}}$
According to question,
$\dfrac{{\Delta \lambda }}{{{\lambda _1}}} = 0.5\% = \dfrac{{0.5}}{{100}}$
Hence, by replacement of known values, we have
$0.005 = \dfrac{P}{{{p_1}}}$
Hence, by making ${p_1}$ the subject of the formula, we have
${p_1} = \dfrac{P}{{0.005}} = 200P$

Hence, the correct option is D.
Note: For clarity, observe that the negative sign was dropped from $d\lambda = - \dfrac{\lambda }{p}dp$ as it became$\dfrac{{\Delta \lambda }}{\lambda } = \dfrac{{\Delta p}}{p}$. This is because, wavelength and momentum are inversely proportional to each other, hence, increase in one causes a decrease in the other, so change in momentum and change in wavelength are negative of each other.
$\Delta \lambda = - \Delta p$