91Ó°ÊÓ

Determine the uncertainty principle formula:

\(\Delta x\Delta p = \frac{h}{{2\pi }}\)

Here,\(\Delta x\)is the uncertainty in the position and\(\Delta p\)is the uncertainty in the momentum.

Consider the expression for the momentum is\[\Delta p = m\Delta v\].

Determine the uncertainty in position as:

\[\begin{array}{c}\Delta x{\rm{ }} = \frac{{6.63 \times {{10}^{ - 34}}{\rm{ J}} \cdot {\rm{s}}}}{{4\pi \times 9.11 \times {{10}^{ - 31}}{\rm{ kg}} \times 2.0 \times {{10}^3}\;{\rm{m}}{{\rm{s}}^{ - 1}}}}\\ = 2.9 \times {10^{ - 8}}{\rm{ m}}\end{array}\]

Compare the uncertainty with the size of the atom as follows:

\[\begin{array}{c}\frac{{\Delta x}}{s} = \frac{{2.9 \times {{10}^{ - 8}}\;{\rm{m}}}}{{0.1 \times {{10}^{ - 9}}}}\\ = 290\end{array}\]

Therefore, the uncertainty in position is 2.9 x 10^-8m. Also, the value is 290 larger than size of the atom.