Question

The de Broglie wavelength $\lambda$of an electron accelerated through a potential V in Volts is:

• A. $\frac{1.227}{\sqrt{V}}nm$
• B. $\frac{0.1227}{\sqrt{V}}nm$
• C. $\frac{0.01227}{\sqrt{V}}nm$
• D. $\frac{0.1227}{\sqrt{V}}nm$

Answer 1

Answer: (A)

$\frac{1.227}{\sqrt{V}}nm$

Answer 2

: Consider an electron of mass m and charge e accelerated from rest through potential V. then

$K = eV$

$K = \frac{1}{2}mv^{2} = \frac{p^{2}}{2m}\therefore p = \sqrt{2mk} = \sqrt{2mev}$

The de Broglie wavelength $\lambda$of the electron is

$\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mk}} = \frac{h}{\sqrt{2meV}}$

Substituting the numerical values of h, m, e, we get

$\lambda = \frac{6.63 \times 10^{- 34}}{\sqrt{2 \times 9.1 \times 10^{- 31} \times 1.6 \times 10^{- 19} \times V}}$

$= \frac{1.227 \times 10^{- 9}}{\sqrt{V}}m = \frac{1.227}{\sqrt{V}}nm$