Question

What is the closest value of de-Broglie wavelength of an electron accelerated by a voltage of 50V? $ (\mid \text{e}\mid =\text{1}.\text{6 }\times \text{ 1}{{0}^{-\text{19}}}\text{C},\text{ }{{\text{m}}_{\text{e}}}=\text{ 9}.\text{1 }\times \text{ 1}{{0}^{-\text{31}}}\text{kg},\text{ h }=\text{ 6}.\text{6 }\times \text{ 1}{{0}^{-\text{34}}} $ Js).
(A) $ 0\cdot \text{5}{\mathop A\limits^\circ} $
(B) $ \text{1}\cdot \text{7}{\mathop A\limits^\circ} $
(C) $ \text{2}\cdot \text{4}{\mathop A\limits^\circ} $
(D) $ \text{1}\cdot \text{2}{\mathop A\limits^\circ} $

Answer 1

Hint: Use the de-Broglie wavelength formula i.e.
 $\Rightarrow \lambda =\dfrac{h}{p} $
Where ‘p’ is the momentum calculated from the kinetic energy of the accelerated electron.
The kinetic energy in the form of potential applied to the electrons is given by
 $\Rightarrow E=\left. \left| e \right. \right|V $

Complete step by step solution
De-Broglie wavelength is given by:
 $\Rightarrow \lambda =\dfrac{h}{p} $
Where ‘h’ is the Planck’s constant
‘p’ is the momentum gained by the electron due to acceleration
As we know that kinetic energy in terms of momentum has the formula:
 $\Rightarrow E=\dfrac{{{p}^{2}}}{2{{m}_{e}}} $
Here, me is the mass of the electron
From the above formula we get;
 $\Rightarrow p=\sqrt{2{{m}_{e}}E} $
Here,
 $\Rightarrow E=\left. \left| e \right. \right|V $
In this |e| is the charge of electron and V is the voltage provided to accelerate the electron
Substituting the energy value in momentum form we get;
 $ \begin{align}
  & p=\sqrt{2{{m}_{e}}\left. \left| e \right. \right|V} \\
 & =\sqrt{2\times 9.1\times {{10}^{-31}}\times 1.6\times {{10}^{-19}}\times 50} \\
 & =\sqrt{1456\times {{10}^{-50}}} \\
 & p=38.157\times {{10}^{-25}}\text{ }kg\text{ }m\text{ }{{s}^{-1}} \\
\end{align} $
Now substituting this momentum value in de-Broglie formula of wavelength i.e.
 $ \begin{align}
  & \lambda =\dfrac{h}{p} \\
 & =\dfrac{6.6\times {{10}^{-34}}}{38.157\times {{10}^{-25}}} \\
 & =0.173\times {{10}^{-9}}m \\
 & \lambda \approx 1.7\times {{10}^{-10}}m \\
 & \lambda \approx 1.7 \mathop A\limits^\circ \\
\end{align} $
Hence, the de-Broglie wavelength is $ \lambda \approx 1.7 \mathop A\limits^\circ $ .
Therefore, option (B) is the correct answer.

Note
Here $ {{m}_{e}} $ is the rest mass of the electron and not the relative mass to avoid the complexity of the question. Also the potential energy of the electron here is taken as zero as the whole of potential energy of the electron is converted into its kinetic energy while being accelerated. Hence, total energy is equal to the kinetic energy of the electron.