Question

Calculate the uncertainty in the velocity of a cricket ball of mass 130 g, if the uncertainty in its
position is of the order of 1.2 A0.

Answer 1

Given:

The uncertainty in the position is of the order of 1.2 A°. Mass is 130 g.

To find:

Uncertainty in velocity?

Calculation:

Applying Heisenberg's Uncertainty Principle:

$\therefore \: \Delta x \times \Delta p = \dfrac{h}{4\pi}$

$\implies \: \Delta x \times(m \Delta v) = \dfrac{h}{4\pi}$

$\implies \: (1.2 \times {10}^{ - 10} ) \times( \dfrac{130}{1000} \times \Delta v) = \dfrac{h}{4\pi}$

$\implies \: (1.2 \times {10}^{ - 10} ) \times( \dfrac{13}{100} \times \Delta v) = \dfrac{h}{4\pi}$

$\implies \: 0.09\times {10}^{ - 12} \times \Delta v = \dfrac{h}{4\pi}$

$\implies \: 9\times {10}^{ - 14} \times \Delta v = \dfrac{h}{4\pi}$

$\implies \: 9\times {10}^{ - 14} \times \Delta v = \dfrac{6.63 \times {10}^{ - 34} }{4\times 3.14}$

$\implies \: 9\times {10}^{ - 14} \times \Delta v = 0.52 \times {10}^{ - 34}$

$\implies \: \Delta v = 0.058\times {10}^{ - 20}$

$\implies \: \Delta v = 5.8\times {10}^{ - 22} \: m/s$

So, uncertainty in velocity is 5.8 × 10^(-22) m/s.

Answer 2

Solution:

By using Heisenberg's Uncertainty Principle,

$\bullet \: {\underline{\boxed{\mathcal{\pmb{\blue{\quad \delta x \times \delta p = \frac{h}{4 \pi} }}}}}}$

Substituting values,

$\leadsto\rm{ \delta x \times (m \delta v) = \frac{h}{4\pi} } \\ \\ \leadsto\rm{(1.2 \times {10}^{ - 10} \times ( \frac{130}{1000} \times \delta v) = \frac{h}{4\pi} } \\ \\ \leadsto\rm{(1.2 \times {10}^{ - 10} \times ( \frac{13}{100} \times \delta v) = \frac{h}{4\pi} } \\ \\ \leadsto\rm{0.09 \times {10}^{ - 12} \times \delta v = \frac{h}{4\pi} } \\ \\ \leadsto\rm{9 \times {10}^{ - 14} \times \delta v = \frac{h}{4\pi} } \\ \\ \leadsto\rm{9 \times {10}^{ - 14} \times \delta v = \frac{6.33 \times {10}^{ - 34} }{4 \times 3.14} } \\ \\ \leadsto\rm{9 \times {10}^{ - 14} \times \delta v = 0.52 \times {10}^{ - 34} } \\ \\ \leadsto\rm{ \delta v = 0.058 \times {10}^{20} }$

════════ ✥.❖.✥ ════════

$\longrightarrow {\mathcal{\sf{\red{\delta v = 5.8 \times {10}^{ - 22} m/s }}}} \: \bigstar$