A baseball has mass $143 \mathrm {~g}$ and a speed of $45 \mathrm {~m} / \mathrm { s }$, with the speed known to within $0.10 \%$. What is the minimum uncertainty in the position of the baseball? $\left( h = 6.626 \times 10 ^ { - 34 } \mathrm {~J} \cdot \mathrm { s } \right)$
A) $1.6 \times 10 ^ { - 32 } \mathrm {~m}$
B) $1.8 \mathrm {~nm}$
C) $1.4 \times 10 ^ { - 32 } \mathrm {~m}$
D) $1.4 \times 10 ^ { - 30 } \mathrm {~m}$
E) $1.6 \times 10 ^ { - 30 } \mathrm {~m}$

Question

Quizplus · Accepted Answer

The uncertainty in the position of the baseball is given by the Heisenberg uncertainty principle:$$\Delta x \Delta p \geq \frac{h}{4\pi}$$where $\Delta x$ is the uncertainty in position, $\Delta p$ is the uncertainty in momentum, and $h$ is the Planck constant.The uncertainty in momentum is given by:$$\Delta p = m \Delta v$$where $m$ is the mass of the baseball and $\Delta v$ is the uncertainty in speed.The uncertainty in speed is given as $0.10\%$ of the speed, or:$$\Delta v = 0.001 \times 45 \mathrm {~m} / \mathrm { s } = 0.045 \mathrm {~m} / \mathrm { s }$$Substituting these values into the Heisenberg uncertainty principle, we get:$$\Delta x \geq \frac{h}{4\pi m \Delta v} = \frac{6.626 \times 10 ^ { - 34 } \mathrm {~J} \cdot \mathrm { s }}{4\pi \times 0.143 \mathrm {~kg} \times 0.045 \mathrm {~m} / \mathrm { s }} = 1.6 \times 10 ^ { - 32 } \mathrm {~m}$$Therefore, the minimum uncertainty in the position of the baseball is $1.6 \times 10 ^ { - 32 } \mathrm {~m}$.

A Baseball Has Mass 143 g143 \mathrm {~g}143 g And a Speed Of 45 m/s45 \mathrm {~m} / \mathrm { s }45 m/s

A Baseball Has Mass $143 \mathrm {~g}$ And a Speed Of $45 \mathrm {~m} / \mathrm { s }$