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Question: If umcertainty principle is applied to an object of mass 1 milligram, the uncertainty value of veloc...

If umcertainty principle is applied to an object of mass 1 milligram, the uncertainty value of velocity and position will be.

A

10−4m2s−110^{- 4}m^{2}s^{- 1}

B

106m2s−110^{6}m^{2}s^{- 1}

C

10−28m2s−110^{- 28}m^{2}s^{- 1}

D

10−34m2s−110^{- 34}m^{2}s^{- 1}

Answer

10−28m2s−110^{- 28}m^{2}s^{- 1}

Explanation

Solution

: Δv.Δx=h4πm=6.626×10−34Js4×3.14×10−6kg=10−28m2s−1\Delta v.\Delta x = \frac{h}{4\pi m} = \frac{6.626 \times 10^{- 34}Js}{4 \times 3.14 \times 10^{- 6}kg} = 10^{- 28}m^{2}s^{- 1}

(1mg=10−6kg)(1mg = 10^{- 6}kg)