An electron has a speed of 30,000 cm sec^-1 accurate up to 0.001%. What is the uncertainty in locating its position in centimeter ? $(m_{e} = 9 \times 10^{- 31} kg) (h = 6.6 \times 10^{- 34} J \sec)$ (round off to nearest integer)

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Detailed Solution

$Δ v = 300 \times 10^{- 5} m / s$
$(Δ x) (m Δ v) = \frac{h}{4 π}$
$(Δ x) = \frac{h}{4 π m Δ v} = \frac{6.6 \times 10^{- 34} \times 7}{4 \times 22 \times 9 \times 10^{- 31} \times 3 \times 10^{- 3}}$

$Δ x = \frac{6.6 \times 100 \times 7}{4 \times 22 \times 9 \times 3} = \frac{70}{36} ≃ 2 cm$

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