Question

De broglie wavelength associated with an electron accelerated through a potential difference of $200v$ is nearly

Answer 1

Step 1:Given:

Voltage, $V=200\hspace{0.17em}\text{v}$.

Step 2:Write an expression for De broglie wavelength as follows:

$\lambda =\frac{12.27}{\sqrt{V}}\dot{A}$

Step 3:Substitute $200\text{v}$for $V$ in expression of De-broglie wavelength:

$$\begin{gathered} \lambda =\frac{12.27}{\sqrt{200}}\stackrel{o}{A} \\ =1.227\stackrel{o}{A} \end{gathered}$$

Hence, De broglie wavelength is $1.227\stackrel{o}{A}$.