Q. The uncertainty in the momentum of an electron is $1.0 \times 10^{- 5}kg\ ms^{- 1}$. The uncertainty in its position will be
A) $1.50 \times 10^{- 28}m$
B) $1.05 \times 10^{- 26}m$
C) $5.27 \times 10^{- 30}m$
D) $5.25 \times 10^{- 28}m$
Explanation
(c)
$\mathrm{\Delta}x \times \mathrm{\Delta}p \geq \frac{h}{4\pi}$
where, $\mathrm{\Delta}x =$uncertainty in position.
$\mathrm{\Delta}p$=uncertainty in momentum.
$= 1.0 \times 10^{- 5}kg\ ms^{- 1}$
$\therefore\ \mathrm{\Delta}x \times 1.0 \times 10^{- 5} \geq \frac{6.62 \times 10^{- 34}}{4 \times 3.14}$
$\mathrm{\Delta}x \geq \frac{6.62 \times 10^{- 34}}{4 \times 3.14 \times 1.0 \times 10^{- 5}}$
$\ \ \ \ \ \ \geq 5.27 \times 10^{- 30}m$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.