Question

An electron is accelerated from rest, between two points A and B at which the potentials are 20V and 40V respectively. The de Broglie wavelength associated with the electron at B will be:
\[\begin{align}
  & \text{A) 0}\text{.75}A{}^\circ \\
 & \text{B) 7}\text{.5}A{}^\circ \\
 & \text{C) 2}\text{.75}A{}^\circ \\
 & \text{D) 2}\text{.75}m \\
\end{align}\]

Answer 1

Hint: De Broglie wavelength is the wavelength associated with a moving particle. de Broglie wavelength is given by the ratio of Planck’s constant and momentum of a particle, as the particle is in motion it will have momentum. Here we will solve the de Broglie relation for an electron to get the de Broglie wavelength associated with the electron.
Formula used:
\[\begin{align}
  & \lambda =\dfrac{h}{p} \\
 & p=\sqrt{2mK.E.} \\
\end{align}\]

Complete answer:
As the light has dual nature of particle and wave, therefore a particle will have a wave characteristic too. de Broglie worked on this thesis and gave a relation between wave and particle. He gave de Broglie wavelength for a moving particle having momentum, say p. The expression for de Broglie wavelength is
\[\lambda =\dfrac{h}{p}\]
Where h is the Planck’s constant and p is the momentum of a particle.
Now the momentum can be given in terms of kinetic energy and the equation for it is
\[p=\sqrt{2mK.E.}\]
Substituting this value of momentum in de Broglie relation, we get
\[\lambda =\dfrac{h}{\sqrt{2mK.E.}}..........\text{(i)}\]
For an electron, K.E. can be given as a product of charge and the potential difference associated with the charge for carrying it from one point to another. Hence we can write K.E. for electron as
\[K.E.=eV\]
Where e is the charge on the electron and V is the potential difference.
Substituting the value of K.E. in equation (i), we have
\[\lambda =\dfrac{h}{\sqrt{2meV}}.............(ii)\]
\[h=6.62\times {{10}^{-34}}Js,\text{ mass of electron }m=9.11\times {{10}^{-31}}kg\text{ and charge on electron }e=1.6\times {{10}^{-19}}C\]
According to the question voltage difference or potential difference is given as
\[\begin{align}
  & V={{V}_{1}}-{{V}_{2}} \\
 & V=40-20 \\
 & V=20V \\
\end{align}\]
Substituting all the values in equation (ii), we have
\[\begin{align}
  & \lambda =\dfrac{6.62\times {{10}^{-34}}}{\sqrt{2\left( 9.11\times {{10}^{-31}} \right)\left( 1.6\times {{10}^{-19}} \right)\left( 20 \right)}} \\
 & \Rightarrow \lambda =\dfrac{6.62\times {{10}^{-34}}}{\sqrt{583.04\times {{10}^{-50}}}} \\
 & \Rightarrow \lambda =\dfrac{6.62\times {{10}^{-34}}}{24.1\times {{10}^{-25}}} \\
 & \Rightarrow \lambda =0.2746\times {{10}^{-9}}m \\
 & \Rightarrow \lambda \simeq 2.75\times {{10}^{-10}}m \\
 & \Rightarrow \lambda =2.75A{}^\circ \\
\end{align}\]

So, the correct answer is “Option C”.

Note:
Conversion of units should be done properly in case the wavelength was asked in nm then the wavelength would be 0.275nm. As there are decimal values so due to rounding off the answer may vary a little and the standard values of Planck’s constant, mass and charge of the electron should be substituted correctly.