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Q. If a $1.00\, g$ body is travelling along the $X$ -axis at $100 \,cm$ $s ^{-1}$ within $1\, cm s ^{-1}$, then uncertainity in its position is

Structure of Atom

Solution:

there is uncertainty with $1 cm s ^{-1}$,

hence, velocity varies between $99 \,cm \,s ^{-1}$ to $101 \,cm \,s ^{-1}$

Thus, $\Delta v$ (uncertainty in velocity) $=101-99=2 \,cm\, s ^{-1}$

$=0.02 \,m\, s ^{-1}$

By Heisenberg's uncertainty principle

$\Delta x \cdot \Delta p \approx \frac{h}{4 \pi} $

$\Delta x \cdot m \Delta v \approx \frac{h}{4 \pi}$

$\therefore $ Uncertainty in position,

$\Delta x \approx \frac{h}{4 \pi m \Delta v}$

$\Delta x=\frac{6.62 \times 10^{-34} \,Js \left( kg \,m ^{2} s ^{-2}\right)}{4 \pi 1.0010^{-3} kg \times 0.02\, ms ^{-1}}$

$=2.63 \times 10^{-30} \,m$