Atomic StructureHard

Question

The mass of a particle is 10–10 g and its diameter is 10−4 cm. If its speed is 10–6 cm/s with 0.0001% uncertainty in measurement, the minimum uncertainty in its position is

Options

A.5.28 × 10–8 m
B.5.28 × 10–7 m
C.5.28 × 10–6 m
D.5.28 × 10–9 m

Solution

$\Delta{x\frac{h}{4\pi m.\Delta v}\frac{6.626 \times 10^{- 34}}{4\pi \times 10^{- 13} \times \frac{0.001}{100} \times 10^{- 8}}^{- 8}\text{ m}}_{\min}$

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