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Q.

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

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a

$5 .2 \times 10^{- 7} m$

b

$5 .2 \times 10^{- 6} m$

c

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

d

$5.2 \times 10^{- 9}$

answer is A.

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Detailed Solution

$\begin{array}{l} m = 10^{- 10} g \Rightarrow 10^{- 13} kg \\ Δv = \frac{0.0001}{100} \times 10^{- 6} \times 10^{- 2} = 10^{- 14} m \sec^{- 1} \\ = Δx \cdot Δp = \frac{h}{4 π} \\ \Rightarrow Δx = \frac{h}{4 π Δp} = \frac{h}{4 π \cdot mΔv} \\ Δx = \frac{6.62 \times 10^{- 34}}{4 \times 3.14 \times 10^{- 13} \times 10^{- 14}} \\ Δx = \frac{6.62}{12.56} \times \frac{10^{- 34}}{10^{- 27}}; Δx = 5.2 \times 10^{- 8} m \end{array}$

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