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

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.