Question
The uncertainty in momentum of an electron is $$1 \times {10^{ - 5}}kg/ms.$$ The uncertainty in its position will be
$$\left( {{\text{Given,}}\,h = 6.62 \times {{10}^{ - 34}}kg\,\,{m^2}/s} \right)$$
A.
$$1.05 \times {10^{ - 28}}m$$
B.
$$1.05 \times {10^{ - 26}}m$$
C.
$$5.27 \times {10^{ - 30}}m$$
D.
$$5.25 \times {10^{ - 28}}m$$
Answer :
$$5.27 \times {10^{ - 30}}m$$
Solution :
According to Heisenberg's uncertainty principle
$$\eqalign{
& \Delta p \times \Delta x \geqslant \frac{h}{{4\pi }} \cr
& {\text{Uncertainty in momentum}} \cr
& \Delta p = 1 \times {10^{ - 5}}kg\,m/s \cr
& 1 \times {10^{ - 5}} \times \Delta x = \frac{{6.62 \times {{10}^{ - 34}}}}{{4 \times \frac{{22}}{7}}}\,\,\left( {{\text{Given}}} \right) \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\Delta x = \frac{{6.62 \times {{10}^{ - 34}} \times 7}}{{1 \times {{10}^{ - 5}} \times 4 \times 22}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 5.265 \times {10^{ - 30}}m \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \approx 5.27 \times {10^{ - 30}}m \cr} $$