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

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;
\[\lambda = \dfrac{h}{p}\] 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.
\[\Delta \lambda = - \Delta p\]