Question

de-Broglie wavelength of an electron accelerated by a voltage of $50 \,V$ is close to
$\left(\left|e\right| = 1.6\times 10^{-19} C, m_{e} = 9.1\times10^{-31} kg, h = 6.6\times10^{-34} Js\right):$

• A. $0.5\, \mathring{A}$
• B. $1.2\, \mathring{A}$
• C. $1.7\, \mathring{A}$
• D. $2.4\, \mathring{A}$

Answer 1

Answer: (C)

$\lambda=\frac{h}{m v}=\frac{h}{\sqrt{2 m q V}}$
For electron
$m=9.1 \times 10^{-31} kg ; q=1.6 \times 10^{-19} C ;$
$h=6.625 \times 10^{-34}\, Js$
$\lambda=\sqrt{\frac{150}{V}} \mathring{A}=\sqrt{\frac{150}{50}} \mathring{A}=1.7\, \mathring{A}$