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NEET UG CHEMISTRY MCQ: STRUCTURE OF ATOM

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Application +4 | -1
The uncertainty in the velocity of a particle of mass 6.626  x  10−28Kg  is  1  x  10−66.626\,\,x\,\,{10^{^{ - 28}}}Kg\,\,is\,\,1\,\,x\,\,{10^{^{ - 6}}} m/s. What is the uncertainty in its position ( in nm )
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Correct option: D
1/4π1/4\pi
Option D (1/4π1/4\pi) is correct because Heisenberg's uncertainty principle states that ΔxΔp≥h4π\Delta x \Delta p \geq \frac{h}{{4\pi}}, where Δp=mΔv\Delta p = m \Delta v. Solving for Δx\Delta x gives Δx=h4πmΔv=6.626×10−344π×6.626×10−28×1×10−6=14π×10−9m=14πnm\Delta x = \frac{h}{{4\pi m \Delta v}} = \frac{{6.626 \times 10^{-34}}}{{4\pi \times 6.626 \times 10^{-28} \times 1 \times 10^{-6}}} = \frac{1}{{4\pi}} \times 10^{-9} m = \frac{1}{{4\pi}} nm.

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