Question

The uncertainty in momentum of a particle is $3.31\times {10}^{-2}Kg$ $mse{c}^{-1}$. Calculate uncertainty in its position.

• A. $6.4\times {10}^{-33}m$
• B. $1.6\times {10}^{-33}m$
• C. $3.2\times {10}^{-33}m$
• D. $5\times {10}^{-33}m$
• E. $2.5\times {10}^{-33}m$

Answer 1

Answer: (B)

$1.6\times {10}^{-33}m$

$\Delta p=3.31\times {10}^{-2}kg$ $m$ $se{c}^{-1}$

Since, $\Delta p\times \Delta X=\frac{h}{4\pi }$

$\therefore \Delta X=\frac{h}{4\pi \cdot \Delta p}=\frac{6.626\times {10}^{-34}}{4\times 3.14\times 3.31\times {10}^{-2}}$

$=1.6\times {10}^{-33}m$

Hence the correct option is B.