Question

How is the de-Broglie wavelength associated with an electron accelerated through a potential difference of 100 volts?

Answer 1

Hint: This can be solved by using ideas like the conservation of mechanical energy because electrostatic force is a conservative force and there is no non conservative force and finding out the de-broglie wavelength when momentum is known.
Formula used:
$\eqalign{
  & PE = eV \cr
  & KE = \dfrac{{{p^2}}}{{2m}} \cr
  & \lambda = \dfrac{h}{p} \cr} $

Complete answer:
Complete solution Step-by-Step:
It is well known that the electron exhibits the particle nature and wave nature. Since an electron is a light particle and it moves with very high velocity it shows wavy nature. We can find out that wavelength by using the formula of de Broglie wavelength. Let us assume de Broglie wavelength is $\lambda $ and $p$ is the momentum of an electron and $h$ is the Planck's constant.
The de Broglie wavelength will be
$\lambda = \dfrac{h}{p}$
When an electron of charge ‘e’ is accelerated through the potential difference ‘V’ then the potential energy possessed by that electron will be $PE = eV$
The kinetic energy of an electron of mass ‘m’ and momentum ‘p’ will be $KE = \dfrac{{{p^2}}}{{2m}}$
Since the electron is acted upon by conservative force only we use conservation of mechanical energy here i.e the initial potential energy is converted to the kinetic energy by the time it reaches the voltage source.

$\eqalign{
  & PE = KE \cr
  & \Rightarrow eV = \dfrac{{{p^2}}}{{2m}} \cr
  & \Rightarrow p = \sqrt {2meV} \cr} $

Hence the de Broglie wavelength will be
$\lambda = \dfrac{h}{p}$
$\eqalign{
  & \Rightarrow \lambda = \dfrac{h}{{\sqrt {2meV} }} \cr
  & \Rightarrow \lambda = \dfrac{{6.6 \times {{10}^{ - 34}}}}{{\sqrt {2(9.1 \times {{10}^{ - 31}})(1.6 \times {{10}^{ - 19}})100} }} \cr
  & \Rightarrow \lambda = 1.22 \times {10^{ - 10}}m \cr} $
We have substituted the values of planck’s constant and mass of electron and charge of electron and potential difference through which electron is accelerated in the above de Broglie wavelength formula and got the value of de Broglie wavelength as $1.22 \times {10^{ - 10}}m$

Note:
For the normal objects which have considerable mass even though they move at higher speeds we can’t consider wavy motion because due to their considerable mass the value of de Broglie wavelength will be too small so that we can neglect it and hence they exhibit only particle nature.