Question

An electron moves in an electric field with a kinetic energy of 2.5 eV. What is the associated de-Broglie wavelength?

Answer 1

Hint: In order to calculate the de-Broglie wavelength, we need to find the relation between the kinetic energy and the de-Broglie wavelength and before that, we need to find the relation between velocity of electron and the wavelength associated to it.

Complete step by step answer:
As we know that the kinetic energy is equal to the half of the product of mass of an electron and the square of the velocity of the electron. Mathematically, this can be written as:
$K.E = \dfrac{1}{2} \times m \times {v^2}....(i)$
Now, from the relation between momentum of the particle and de-Broglie wavelength, we know that:
$p = \dfrac{h}{\lambda }….(ii)$
Also, $p = m \times v....(iii)$
Thus, equating (ii) and (iii), we have:
$
  m \times v = \dfrac{h}{\lambda } \\
  v = \dfrac{h}{{m\lambda }}....(iv) \\
$
Now, equating (iv) with (i), we have:
$
  K.E = \dfrac{1}{2} \times m \times {\left( {\dfrac{h}{{m\lambda }}} \right)^2} \\
  \lambda = \dfrac{h}{{\sqrt {2 \times m \times K.E} }} \\
$
Where, $lambda$ = de-Broglie wavelength
m = mass of an electron = $9.1 \times {10^{ - 31}}kg$
K.E = kinetic energy of electron = 2.5 eV = $2.5 \times 1.6 \times {10^{ - 19}}J$
h = Planck’s constant = $6.626 \times {10^{ - 34}}Js$
Now, substituting the values and solving, we have:
$
  \lambda = \dfrac{{6.626 \times {{10}^{ - 34}}}}{{\sqrt {2 \times 9.1 \times {{10}^{ - 31}} \times 2.5 \times 1.6 \times {{10}^{ - 19}}} }} \\
  \lambda = 7.7 \times {10^{ - 10}}m = 7.7\mathop A\limits^o \\
$
Thus, the de-Broglie wavelength associated with the electron is equal to\[7.7 \times {10^{ - 10}}m\].

Note:
Whenever a particle has a wavelength equal to de-Broglie wavelength, all the energy gets converted into kinetic energy and the particle is completely in motion. The greater the de-Broglie wavelength, the lower the energy is. Thus, the lower will be the momentum of the particle.