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

Hint: 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.