Question

If the kinetic energy of a free electron is made double; the new De-Broglie wavelength will be _______ times that of the initial wavelength.
(A) $\dfrac{1}{{\sqrt 2 }}$
(B) $\sqrt 2 $
(C) $2$
(D) $\dfrac{1}{2}$

Answer 1

Hint: The kinetic energy and the De-Broglie wavelength of a free electron are related as \[\lambda = \dfrac{h}{{\sqrt {2mK} }}\] . So in the two cases by taking the ratio of the wavelengths, we can find the number of times the final wavelength increases when the kinetic energy becomes double.

Formula Used: In the solution to this problem, we will be using the following formula,
\[\lambda = \dfrac{h}{{\sqrt {2mK} }}\]
where \[\lambda \] is the De-Broglie wavelength
\[h\] is the Planck’s constant
\[m\] is the mass of the electron
and \[K\] is the kinetic energy.

Complete Step by Step Solution: The De-Broglie wavelength of any object is given by the equation,
\[\lambda = \dfrac{h}{p}\] where \[p\] is the momentum of the body.
Now according to the question, we need to find the De-Broglie wavelength of an electron. So if we consider the velocity of the electron as \[v\], then the momentum is given by, \[p = mv\].
Substituting this in the equation for De-Broglie wavelength we get,
\[\lambda = \dfrac{h}{{mv}}\]
The kinetic energy of an electron moving with a velocity \[v\] is given by,
\[K = \dfrac{1}{2}m{v^2}\]
From here we can find the velocity in terms of the kinetic energy as,
\[{v^2} = \dfrac{{2K}}{m}\]
On taking square root on both sides we get,
\[\Rightarrow v = \sqrt {\dfrac{{2K}}{m}} \]
We substitute this value in the equation of the De-Broglie wavelength. Therefore, we get
\[\Rightarrow \lambda = \dfrac{h}{{m\sqrt {\dfrac{{2K}}{m}} }}\]
On cancelling the mass \[m\] we get
\[\Rightarrow \lambda = \dfrac{h}{{\sqrt {2mK} }}\]
So in the first case, the wavelength is \[{\lambda _1}\] and the kinetic energy is \[{K_1}\]. Therefore,
\[\Rightarrow {\lambda _1} = \dfrac{h}{{\sqrt {2m{K_1}} }}\]
For the second case, wavelength is \[{\lambda _2}\] and the kinetic energy according to the question is \[\Rightarrow {K_2} = 2{K_1}\]
So we get
\[\Rightarrow {\lambda _2} = \dfrac{h}{{\sqrt {2m{K_2}} }} = \dfrac{h}{{\sqrt {2m \times 2{K_1}} }}\]
we can write this as
\[\Rightarrow {\lambda _2} = \dfrac{h}{{\sqrt 2 \times \sqrt {2m{K_1}} }}\]
Now we have already calculated \[{\lambda _1} = \dfrac{h}{{\sqrt {2m{K_1}} }}\]. So substituting this in the equation of \[{\lambda _2}\] gives us
\[\Rightarrow {\lambda _2} = \dfrac{1}{{\sqrt 2 }}{\lambda _1}\]
Therefore the new De-Broglie wavelength will be \[\dfrac{1}{{\sqrt 2 }}\] times the initial wavelength.

So the correct answer will be option A. \[\dfrac{1}{{\sqrt 2 }}\]

Note: According to the wave particle duality in quantum mechanics, it is theorised that not only light but every object has a wave nature. The De-Broglie wavelength is the probability of finding an object in a given point and is inversely proportional to the momentum of the body.