Q. What accelerating potential is needed to produce an electron beam with an effective wavelength of 0.090Å?
A) $1.86 \times 10^{4}eV$
B) $1.86 \times 10^{2}eV$
C) $2.86 \times 10^{4}eV$
D) $2.86 \times 10^{2}eV$
Explanation
(a)
$\lambda = \frac{h}{\sqrt{2m(KE)}}$
$KE = \frac{h^{2}}{2\lambda^{2}m}$
$= \frac{\left( 6.626 \times 10^{- 34} \right)^{2}}{2 \times \left( 0.090 \times 20^{- 10} \right)^{2} \times 9.1 \times 10^{- 31}}$
$= 2.98 \times 10^{- 15}J$
Accelerating potential
$= \frac{2.98 \times 10^{- 15}}{1.6 \times 10^{- 19}}\ eV$
$= 1.86 \times 10^{4}eV$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.