Question

Calculate kinetic energy of an electron having wavelength 1nm.
A. 2.1 eV
B. 3.1 eV
C. 1.5 eV
D. 4.2 eV

Answer 1

Hint: We will use the equation to find de-Broglie wavelength to find the kinetic energy of the electron with wavelength 1nm. 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 $K.E=\dfrac{{{P}^{2}}}{2m}$.
Formula used:
$\lambda =\dfrac{h}{\sqrt{2m{{E}_{kinetic}}}}$

Complete 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.
      And P is the momentum of the particle.
But we know that, momentum of a particle is related to its kinetic energy as,
$K.E=\dfrac{{{P}^{2}}}{2m}$
Where, m is the mass of that particle.
Now,
$\begin{align}
  & K.E=\dfrac{{{P}^{2}}}{2m} \\
 & \Rightarrow P=\sqrt{2m\left( K.E \right)} \\
\end{align}$
Now, substituting this equation for P in the de-Broglie equation, we will get,
$\lambda =\dfrac{h}{\sqrt{2m\left( K.E \right)}}$
Now, to find the wavelength of the electron with wavelength 1 nm, we will use this equation.
$\lambda =\dfrac{h}{\sqrt{2m\left( K.E \right)}}$
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,
$\begin{align}
  & \lambda =\dfrac{h}{\sqrt{2m\left( K.E \right)}} \\
 & \Rightarrow K.E=\dfrac{{{h}^{2}}}{2m{{\lambda }^{2}}} \\
 & \Rightarrow K.E=\dfrac{{{\left( 6.626\times {{10}^{-34}} \right)}^{2}}}{2\left( 9.1\times {{10}^{-31}}kg \right){{\left( 1\times {{10}^{-9}} \right)}^{2}}} \\
 & \Rightarrow K.E=2.42\times {{10}^{-19}}J \\
\end{align}$
To convert the unit joule into eV, we will further divide it with the charge of an electron.
$\Rightarrow K.E=\dfrac{2.42\times {{10}^{-19}}}{1.6\times {{10}^{-19}}}\approx 1.5eV$
Therefore, kinetic energy of an electron with wavelength 1nm is found to be 1.5eV.

So, the correct answer is “Option C”.

Note:
In questions like this, remembering the relation between wavelength and kinetic energy of a particle directly will help to solve the problem quickly. Also, after finding the kinetic energy in S.I units and converting it to eV is important as options are given in that unit. Joule can be converted into electron volts by dividing it with the modulus of charge of an electron. i.e. $1.6\times {{10}^{-19}}$.