Question

The de Broglie wavelength of an electron moving with kinetic energy of $144 \,eV$ is nearly

• A. $102 \times 10^{-3}\,nm$
• B. $102 \times 10^{-4}\,nm$
• C. $102 \times 10^{-5}\,nm$
• D. $102 \times 10^{-2}\,nm$

Answer 1

Answer: (A)

$\lambda=\frac{12.27}{\sqrt{V}} \mathring{A}$
$=\frac{12.27}{\sqrt{144}}\times10^{10}$
$=1.02\times10^{-10}\,m$
$=102\times10^{-3}\,nm$