Q. An electron beam is accelerated through a potential difference of 10,000 volt. The de-Broglie wavelength of the electron beam is
A) $0.123\ A{^\circ}$
B) 0.356 $A{^\circ}$
C) 0.186$\ A{^\circ}$
D) 0.258$\ A{^\circ}$
Explanation
(a)
$\lambda = \frac{h}{\sqrt{2eVm_{e}}}$
$e = 1.6 \times 10^{- 19}C,V = 10,000\ V,m_{e} = 9.1 \times 10^{- 31}$ kg
$\lambda = \frac{6.63 \times 10^{- 34}}{\sqrt{2 \times 1.6 \times 10^{- 19} \times 10,000 \times 9.1 \times 10^{- 31}}} = 0.123\ \mathring{\mathrm{A}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.