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Chemistry
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
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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
A
$1.318 \times 10^{-30} \,m$
B
$5.272 \times 10^{-30} \,m$
C
$2.636 \times 10^{-30}\, m$
D
$6.590 \times 10^{-31}\, m$
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,
$\Delta x \approx \frac{h}{4 \pi m \Delta v}$
$\Delta x=\frac{6.62 \times 10^{-34} \,Js \left( kg \,m ^{2} s ^{-2}\right)}{4 \pi 1.0010^{-3} kg \times 0.02\, ms ^{-1}}$
$=2.63 \times 10^{-30} \,m$