Question

The de-Broglie wavelength of a free electron with kinetic energy $'E'$ is $\lambda$. If the kinetic energy of the electron is doubled, the de-Broglie wavelength is
$A.\dfrac{\lambda }{{\sqrt 2 }}$
$B.\sqrt 2 \lambda$
$C.\dfrac{\lambda }{2}$
$D.2\lambda$

Answer 1

We will use the equation to find de-Broglie wavelength to find the kinetic energy of the electron with wavelength $\lambda$. De-Broglie wavelength of a particle is inversely proportional to the momentum of that particular body. We should know that kinetic energy and momentum of a particle is related as $K.E = \dfrac{{{P^2}}}{{2m}}$ where $P$ is momentum of body and $m$ is the mass of the body.

Formula used:
$\lambda = \dfrac{h}{{\sqrt {2mK.E} }}$

Complete step by step answer:
The formula for finding de-Broglie wavelength is given as,
$\lambda = \dfrac{h}{P} = \dfrac{h}{{mv}}$
Where, $h$ is the planck's constant.
Where, $m$ is the mass of that particle.
Now, $K.E = \dfrac{{{P^2}}}{{2m}}$
$\Rightarrow P = \sqrt {2m\left( {K.E} \right)}$
Now, substituting this equation for P in the de-Broglie equation, we will get,
$\lambda = \dfrac{h}{P} = \dfrac{h}{{\sqrt {2m\left( {K.E} \right)} }}$
Where, $\lambda$ is the wavelength of the electron.
$h$ is the planck's constant having the value $6.626 \times {10^{ - 34}}$.
$m$ is the mass of the electron) $9.1 \times {10^{ - 31}}kg$.
And K.E is the kinetic energy of the electron.
So,the de-Broglie wavelength of a free electron with kinetic energy $K.E$ is given as $\lambda$.Now we find the de-Broglie wavelength if the kinetic energy gets doubled
${\lambda _1} = \lambda$
$K.{E_1} = K.E$
$K.{E_2} = 2K.E$
${\lambda _2} = ?$
From the above derived formulae we can write
${\lambda _1} = \dfrac{h}{{\sqrt {2m\left( {K.{E_1}} \right)} }}$ …….(1)
${\lambda _2} = \dfrac{h}{{\sqrt {2m\left( {K.{E_2}} \right)} }}$……..(2)
Now we take ratio of (1) by (2) we get
$\dfrac{{{\lambda _1}}}{{{\lambda _2}}} = \dfrac{{\dfrac{h}{{\sqrt {2m\left( {K.{E_1}} \right)} }}}}{{\dfrac{h}{{\sqrt {2m\left( {K.{E_2}} \right)} }}}}$
$\Rightarrow \dfrac{{{\lambda _1}}}{{{\lambda _2}}} = \dfrac{{\sqrt {2m\left( {K.{E_2}} \right)} }}{{\sqrt {2m\left( {K.{E_1}} \right)} }} = \dfrac{{\sqrt {\left( {K.{E_2}} \right)} }}{{\sqrt {\left( {K.{E_1}} \right)} }} = \sqrt {\dfrac{{\left( {K.{E_2}} \right)}}{{\left( {K.{E_1}} \right)}}}$
We were given the kinetic energy get doubled so taking $K.{E_2} = 2K.{E_1}$ we get
$\dfrac{{{\lambda _1}}}{{{\lambda _2}}} = \sqrt {\dfrac{{\left( {2K.{E_1}} \right)}}{{\left( {K.{E_1}} \right)}}} = \sqrt 2$
$\Rightarrow {\lambda _2} = \dfrac{{{\lambda _1}}}{{\sqrt 2 }}$
Therefore, the de-Broglie wavelength becomes $\dfrac{\lambda }{{\sqrt 2 }}$ if the kinetic energy gets doubled.

Note:
In questions like this, remembering the relation between wavelength and kinetic energy of a particle directly will help to solve the problem quickly. The matter shows wave- particle duality relation that means at one moment the matter shows the wave behaviour and at another moment it acts like a particle. De-Broglie used the wave behaviour of the matter and gave the relations between the momentum and the wavelength of the particle.