Question

An electron of mass $9.1\times {10}^{-31}$kg and charge $1.6\times {10}^{-19}$C is accelerated through a potential difference of 'V' volt. The de Broglie wavelength $\left(\lambda \right)$ associated with the electron is

• A. $\frac{12.27}{\sqrt{V}}{A}^0$
• B. $\frac{12.27}{V}{A}^0$
• C. $12.27\sqrt{V}{A}^0$
• D. $\frac{1}{12.27\sqrt{V}}{A}^0$

Answer 1

Answer: (A)

$\frac{12.27}{\sqrt{V}}{A}^0$

$\frac{1}{2}m{v}^2=eV$
$mv=\sqrt{2meV}$
Now $\lambda =\frac{h}{mv}$
$=\frac{h}{\sqrt{2meV}}$
$=\frac{12.27}{\sqrt{V}}{A}^0$