Question

The de Broglie wavelength A 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. $\frac{0.001227}{\sqrt{V}}nm$
• D. $\frac{12.27}{\sqrt{V}}nm$

Answer 1

Answer: (A)

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

Consider an electron of mass m and charge e accelerated from rest through potential V. Then, K = eV

$K=\frac{1}{2}m{v}^2=\frac{p^2}{2m}$

$\therefore$ $p=\sqrt{2mK}=\sqrt{2meV}$

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

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

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

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

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