Q. If the mass of neutral $= 1.7 \times 10^{- 27}$kg, then the de-Broglie wavelength of neutral of energy 3eV is $(h = 6.6 \times 10^{- 34}J - s)$
A) $1.6 \times 10^{- 16}$m
B) $1.6 \times 10^{- 11}$m
C) $1.4 \times 10^{- 10}$m
D) $1.4 \times 10^{- 11}$m
Explanation
(b) $E = 3eV = 3 \times 1.6 \times 10^{- 19}$ J
$\lambda = \frac{h}{\sqrt{2mE}}$
$= \frac{6.6 \times 10^{- 34}}{\sqrt{2 \times 1.7 \times 10^{- 27} \times 3 \times 1.6 \times 10^{- 19}}} = 1.65 \times 10^{- 11}m$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.