Question

The de Broglie wavelength of an electron with kinetic energy $120\,eV$ is (Given $h=6.63\times10^{-34}\,J\,s, m_{e}=9\times10^{-31}\,kg, 1\,e\,V =1.6\times10^{-19}\,J)$

• A. $2.13\,\mathring{A}$
• B. $1.13\,\mathring{A}$
• C. $4.15\,\mathring{A}$
• D. $3.14\,\mathring{A}$

Answer 1

Answer: (B)

As $p=\sqrt{2mK}$
$=\sqrt{2\times9\times10^{-31}\times120\times1.6\times10^{-19}}$
$=5.88\times10^{-24}\,kg\,m\,s^{-1}$
de Broglie wavelength,
$\lambda=\frac{h}{p}=\frac{6.63\times10^{-34}}{5.88\times10^{-24}}=1.13\times10^{-10}m$
$=1.13 \,\mathring{A}$