Question

What is the de Broglie wavelength of an electron with a kinetic energy of $120\text{eV}$?
(Given: $h=6.63\times {10}^{-34}\text{Js},m_e=9.11\times {10}^{-31}\text{kg},e=1.6\times {10}^{-19}\text{coulomb}$

• A. $725\text{pm}$
• B. $500\text{pm}$
• C. $322\text{pm}$
• D. $112\text{pm}$

Answer 1

The correct option is D $112\text{pm}$

For a particle, de Broglie wavelength $\lambda =\frac{h}{p}$
Kinetic energy $K=\frac{p^2}{2m}$
$\therefore \lambda =\frac{h}{\sqrt{2mK}}$
For an electron $\lambda =\frac{h}{\sqrt{2m_e{K}_e}}$
$\lambda =\frac{(6.63\times {10}^{-34}J)}{\sqrt{2\times (9.11\times {10}^{-31}\text{kg})\times (120\times 1.6\times {10}^{-9}J)}}$
$\lambda =112\times {10}^{-12}m=112\text{pm}[\therefore 1\text{pm}={10}^{-12}m]$