Question

Find the de-Broglie wavelength of an electron with a kinetic energy of $120eV$.

• A. 95 pm
• B. 102 pm
• C. 112 pm
• D. 124 pm

Answer 1

Answer: (C)

The de-Broglie wavelength of a electron of $m_e$ and kinetic energy $K$ is given by
$\lambda =\frac{h}{\sqrt{2m_{e}K}}$
Here, $m_e=9.1\times 10^{-31}kg$
and $K=120eV=120\times 1.6\times 10^{-19}J$
$\therefore \lambda =\frac{6.6\times 10^{-34}}{\sqrt{2\times 9.1\times 10^{-31}\times 120\times 1.6\times 10^{-19}}}$
$=\frac{6.6\times 10^{-34}}{\sqrt{3494.4\times 10^{-50}}}$
$\lambda =\frac{6.6\times 10^{-34}}{59.1\times 10^{-25}}$
$\lambda =112.0\times 10^{-12}m$
$\lambda =112pm$