The mass of a particle is $10^{- 10} g$ and its radius is $2 \times 10^{- 4} cm .$ If its velocity is $10^{- 6} {cmsec}^{- 1}$ with 0.0001% uncertainty in measurement, the uncertainty in its position is

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$5.2 \times 10^{- 8} m$

$5.2 \times 10^{- 6} m$

$5.2 \times 10^{- 9} m$

$5.2 \times 10^{- 7} m$

answer is A.

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Detailed Solution

$\begin{matrix} Δx . Δp \geq \frac{h}{4 π} \\ Δp = mΔv \\ m = 10^{- 13} kg, v = 10^{- 8} m / s \end{matrix}$

$\begin{matrix} Δv = \frac{0.0001}{100} \times 10^{- 8} = 10^{- 14} m / s \\ Δx = \frac{h}{4 πΔp} = \frac{h}{4 π \times m \times ∆ v} \\ = \frac{6.6 \times 10^{- 34} J . s}{4 \times 3.14 \times 10^{- 13} kg \times 10^{- 14} m / s} = 5.2 \times 10^{- 8} m \end{matrix}$