Question

An electron is accelerated through a voltage of $40kV$. What will be its wavelength?

• A. $0.061\stackrel{\circ }{A}$
• B. $0.011\stackrel{\circ }{A}$
• C. $0.021\stackrel{\circ }{A}$
• D. $0.161\stackrel{\circ }{A}$

Answer 1

The correct option is A

$0.061\stackrel{\circ }{A}$

Step 1. Given data:

$V=40kV$, Voltage by which electron is accelerated.

$V=40\times {10}^{3}V$

$\lambda =?$ , Wavelength associated.

Step 2. Calculating wavelength of the electron,

From De-Broglie equation,

$\lambda =\frac{1.227}{\sqrt{V}}nm$ $\begin{array}{rcl}& & \left[\left(innm\right),1nm={10}^{-9}m\right]\end{array}$

$\begin{array}{rcl}\lambda & =& \frac{1.227}{\sqrt{40\times {10}^3}}nm\\ & \Rightarrow & \lambda =\frac{1.227}{200}\times {10}^{-9}m\\ & \Rightarrow & \lambda =0.0061\times {10}^{-9}m\\ & \Rightarrow & \lambda =0.061\times {10}^{-10}m\end{array}$

$\begin{array}{rcl}\therefore \lambda & =& 0.061\stackrel{\circ }{A}\end{array}$ $\begin{array}{rcl}& & \left[1\stackrel{\circ }{A}={10}^{-10}m\right]\end{array}$

The wavelength associated with the electron is $\begin{array}{rcl}\mathit{\lambda }\mathbf{}& \mathbf{=}& \mathbf{}\mathbf{0}\mathbf{.}\mathbf{061}\mathbf{}\stackrel{\mathbf{\circ }}{\mathbf{A}}\end{array}$

Hence, correct option is (A).