Question

De-Broglie wavelength of an electron accelerated by a voltage of $50 \, \text{V}$ is close to $\left(\left|e\right| = 1.6 \times \left(10\right)^{- 19} \, C , \, m_{e} = 9.1 \times \left(10\right)^{- 31} \, k g , \, \, h = 6.6 \times \left(10\right)^{- 34} \, J \, s\right)$

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

Answer 1

Answer: (C)

De-Broglie wavelength $\lambda $ in given by
$\lambda =\frac{h}{m v}=\frac{h}{\sqrt{2 m e V}}$
and for electron $\lambda =\frac{12.27 \overset{^\circ }{A}}{\sqrt{V}}$
$=\frac{12.27 \overset{^\circ }{A}}{\sqrt{50}}$
$=1.7$ $\overset{^\circ }{A}$