Question

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

• 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
$\lambda=\frac{h}{\sqrt{2 m_{e} K}}$
Here, $m_{e}=9.1 \times 10^{-31} kg $
and $K=120 eV =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}} $
$\lambda=112.0 \times 10^{-12} m$
$\lambda=112 \,pm$