Question

The de-Broglie wavelength of an electron of kinetic energy $9 eV$ is (take, $h=4 \times 10^{-15} eV - s , c=3 \times 10^{10} cm / s$ and the
mass $m_{e}$ of electron as $\left.m_{e} c^{2}=0.5 MeV \right)$

• A. $4 \times 10^{-8} \,cm$
• B. $3 \times 10^{-8} \,cm$
• C. $4 \times 10^{-7} \,cm$
• D. $3 \times 10^{-7}\, cm$

Answer 1

Answer: (A)

For electron, $p c=\sqrt{2 KE \,m_{0} \,c^2}$
$\Rightarrow p c= \sqrt{2 \times 9 eV \times 0.58 \times 10^{6}} $
$\Rightarrow p c=3 \times 10^{3} eV $
Now, de-Broglie wavelength,
$\lambda=\frac{h c}{p c}=\frac{4 \times 10^{-15} \times 3 \times 10^{10}}{3 \times 10^{3}}=4 \times 10^{-8} \,cm$
$(\because c=3 \times 10^{10}\, cm / s )$