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Q. Calculate the minimum uncertainty in velocity of a particle of mass $1.1 \times 10^{-27}$ kg if uncertainty in its position is $3 \times 10^{-10}$ cm.

Structure of Atom

Solution:

$\Delta x. \Delta\ge \frac{h}{4\pi m} $
$ \Delta v\ge\frac{h}{\Delta x.4\pi.m}$
$=\frac{6.6\times10^{-34}}{3\times10^{-10}\times10^{-2}m\times4\times3.14\times1.1\times10^{-27}}$
$=1.5\times10^{-4} m$