Question

An electron is accelerated through a potential difference of $10000\text{V}$. The de Broglie wavelength associated with electron will be

• A. $0.286\stackrel{\circ }{\text{A}}$
• B. $12.27\stackrel{\circ }{\text{A}}$
• C. $1.227\stackrel{\circ }{\text{A}}$
• D. $0.1227\stackrel{\circ }{\text{A}}$

Answer 1

The correct option is D $0.1227\stackrel{\circ }{\text{A}}$

Since, wavelength associated with the electrons is given by,

$\lambda =\frac{12.27}{\sqrt{V}}\stackrel{\circ }{\text{A}}$

$=12.27\times {10}^{-2}\stackrel{\circ }{\text{A}}$

$=0.1227\stackrel{\circ }{\text{A}}$