Q. In an atom, an electron is moving with a speed of 600 m/s with an accuracy of 0.005%. Certainity with which the position of the electron can be located is $\left( \begin{array}{r} h = 6.6 \times 10^{- 34}kg\ m^{2}s^{- 1},\ \\ mass\ of\ electron,\ e_{m} = 9.1 \times 10^{- 31}kg \end{array} \right)$
A) $1.52 \times 10^{- 4}m$
B) $5.10 \times 10^{- 3}m$
C) $1.92 \times 10^{- 3}m$
D) $3.84 \times 10^{- 3}m$
Explanation
(c)
By Heisenberg’s uncertainty principle
$\mathrm{\Delta}x.m\mathrm{\Delta}V = \frac{h}{4\pi}$
$\mathrm{\Delta}V = 0.005\%\ or\$600m/s$\ = \frac{600 \times 0.005}{100} = 0.03$
$\mathrm{\Delta}x \times 9.1 \times 10^{- 31} \times 0.03 = \frac{6.6 \times 10^{- 34}}{4 \times 3.14}$
$Hence,\ \mathrm{\Delta}x = \frac{6.6 \times 10^{- 34}}{4 \times 3.14 \times 0.03 \times 9.1 \times 10^{- 31}}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 1.92 \times 10^{- 3}m$.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.