Question

An electron has a speed of 40 m/s, accurate up to 99.99%. What is the uncertainity in locating its position?

• A. 0.0144 m
• B. 0.0234 m
• C. 0.0567 m
• D. 0.0198 m

Answer 1

The correct option is A 0.0144 m

From Heisenberg uncertainty principle,

$\Delta x\times \Delta p=\frac{h}{4\pi }$

$\Delta x=\frac{h}{4\pi \times m\Delta v}$

where,

$\Delta v=40\times 0.0001$

$m=9.1\times {10}^{-}31Kg$

$h=6.63\times {10}^{-34}J.sec$

$\Delta x=0.0144m$

Hence, option $A$ is correct.