Question

If the momentum of an electron is changed by $$P$$, then the de Broglie wavelength associated with it changes by $$0.5\mathrm{\%}$$. The initial momentum of the electron will be:
(A) 400P
(B) $${\displaystyle \frac{P}{200}}$$
(C) 100P
(D) 200P

Answer 1

Hint: 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;
$$\lambda ={\displaystyle \frac{h}{p}}$$ 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 $$0.5\mathrm{\%}$$. We are to find the initial momentum of the electron.
Now, we recall that the equation of de Broglie can be given as
$$\lambda ={\displaystyle \frac{h}{p}}$$ 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
$${\lambda }_1={\displaystyle \frac{h}{p_1}}$$
For instantaneous change in the wavelength, we differentiate the de Broglie equation, as in
$${\displaystyle \frac{d\lambda }{dp}}=-{\displaystyle \frac{h}{p^2}}$$
$$\Rightarrow d\lambda =-{\displaystyle \frac{h}{p^2}}dp$$
But $$\lambda ={\displaystyle \frac{h}{p}}$$
Hence, $$d\lambda =-{\displaystyle \frac{\lambda }{p}}dp$$
Hence, $${\displaystyle \frac{\mathrm{\Delta }\lambda }{\lambda }}={\displaystyle \frac{\mathrm{\Delta }p}{p}}$$
Then by replacement with initial wavelength and momentum, we have
$${\displaystyle \frac{\mathrm{\Delta }\lambda }{{\lambda }_1}}={\displaystyle \frac{\mathrm{\Delta }p}{p_1}}$$
According to question,
$${\displaystyle \frac{\mathrm{\Delta }\lambda }{{\lambda }_1}}=0.5\mathrm{\%}={\displaystyle \frac{0.5}{100}}$$
Hence, by replacement of known values, we have
$$0.005={\displaystyle \frac{P}{p_1}}$$
Hence, by making $$p_1$$ the subject of the formula, we have
$$p_1={\displaystyle \frac{P}{0.005}}=200P$$

Hence, the correct option is D.
Note: For clarity, observe that the negative sign was dropped from $$d\lambda =-{\displaystyle \frac{\lambda }{p}}dp$$ as it became$${\displaystyle \frac{\mathrm{\Delta }\lambda }{\lambda }}={\displaystyle \frac{\mathrm{\Delta }p}{p}}$$. 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.
$$\mathrm{\Delta }\lambda =-\mathrm{\Delta }p$$