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The size of a mircorscopic particle is 1 micron and its mass is 6 × 10^–13g. If its position may be measured to within 0.1% of its size, the uncertainity in velocity (in cm^–1) is approximately

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10^–8

$\frac{10^{- 7}}{4 π}$

10^–5

$\frac{10^{- 5}}{4 π}$

answer is A.

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Detailed Solution

$\begin{array}{l} Δx = \frac{0 .1}{100} = 10^{- 4} cm; Δx \cdot mΔv \geq \frac{h}{4 π} \\ ΔV = \frac{h}{4 πm} \times \frac{1}{Δx} = \frac{10^{- 7}}{4 π} \end{array}$

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