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Question: The uncertainty in momentum of an electron is \(1 \times 10^{- 5}kg - m/s\). The uncertainty in its ...

The uncertainty in momentum of an electron is 1×10−5kg−m/s1 \times 10^{- 5}kg - m/s. The uncertainty in its position will be

(h=6.62×10−34kg−m2/sh = 6.62 \times 10^{- 34}kg - m^{2}/s)

A

1.05×10−28m1.05 \times 10^{- 28}m

B

1.05×10−26m1.05 \times 10^{- 26}m

C

5.27×10−30m5.27 \times 10^{- 30}m

D

5.25×10−28m5.25 \times 10^{- 28}m

Answer

5.27×10−30m5.27 \times 10^{- 30}m

Explanation

Solution

According to Δx×Δp=h4π\Delta x \times \Delta p = \frac{h}{4\pi}

Δx=hΔp×4π=6.62×10−341×10−5×4×3.14=5.27×10−30m\Delta x = \frac{h}{\Delta p \times 4\pi} = \frac{6.62 \times 10^{- 34}}{1 \times 10^{- 5} \times 4 \times 3.14} = 5.27 \times 10^{- 30}m.