Q. The uncertainty in momentum of an electron is $1 \times 10^{- 5}$ kg m/s. the uncertainty in its position will be $(h = 6.62 \times 10^{- 34}$ kg $m^{2}/s)$
A) 2.36$\times 10^{- 28}$m
B) 5.25$\times 10^{- 28}$m
C) 2.27$\times 10^{- 30}$m
D) 5.27$\times 10^{- 30}$m
Explanation
(d)
$\mathrm{\Delta}x \bullet \mathrm{\Delta}P = \frac{h}{4\pi}$
$\mathrm{\Delta}x = \frac{6.63 \times 10^{- 34}}{4 \times 3.14 \times 10^{- 5}}$
$= \frac{5.27 \times 10^{- 35}}{1 \times 10^{- 5}}$
$= 5.27 \times 10^{- 30}\$m
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.