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

The uncertainty in momentum of an electron is 1×10−5kgm/s1 \times 10^{- 5}kgm/s. The uncertainity in its position will be (h=6.63×10−34Js)(h = 6.63 \times 10^{- 34}Js)

A

5.28×10−30m5.28 \times 10^{- 30}m

B

5.25×10−28m5.25 \times 10^{- 28}m

C

1.05×10−26m1.05 \times 10^{- 26}m

D

2.715×10−30m2.715 \times 10^{- 30}m

Answer

5.28×10−30m5.28 \times 10^{- 30}m

Explanation

Solution

Uncertainity in position Δx=h4π×Δp\Delta x = \frac{h}{4\pi \times \Delta p}

=6.63×10−344×3.14×(1×10−5)=5.28×10−30m= \frac{6.63 \times 10^{- 34}}{4 \times 3.14 \times (1 \times 10^{- 5})} = 5.28 \times 10^{- 30}m.