Calculate the uncertainty in the velocity of a cricket ball of mass 15

1.	Calculate the uncertainty in the velocity of a cricket ball of mass 150 g, if uncertainty in its position is of the order of 1 Å .
Answer» Given, m = 150 g = \(\frac{150}{1000}\)kg = 1.5 x 10^-2 kg Uncertainty in position (∆x) = 1Å = 1 × 10⁻¹⁰m Uncertainty in velocity (∆v) = ? According to Heisenberg uncertainty principle, ∆x × ∆p = \(\frac{h}{4\pi}\) ∆x × m(∆v) = \(\frac{h}{4\pi}\) \(\therefore\) ∆v = \(\frac{h}{4\pi m \times\Delta \mathrm{x}}\) = \(\frac{6.626\times10^{-36}J.sec}{4\times3.14\times1.5\times10^{-2}kg\times1\times10^{-10}m}\) = 3.5 × 10⁻²⁴ m/s.

Calculate the uncertainty in the velocity of a cricket ball of mass 150 g, if uncertainty in its position is of the order of 1 Å .

Answer»

Given, m = 150 g = \(\frac{150}{1000}\)kg = 1.5 x 10^-2 kg

Uncertainty in position (∆x) = 1Å = 1 × 10⁻¹⁰m

Uncertainty in velocity (∆v) = ?

According to Heisenberg uncertainty principle,

∆x × ∆p = \(\frac{h}{4\pi}\)

∆x × m(∆v) = \(\frac{h}{4\pi}\)

\(\therefore\) ∆v = \(\frac{h}{4\pi m \times\Delta \mathrm{x}}\)

= \(\frac{6.626\times10^{-36}J.sec}{4\times3.14\times1.5\times10^{-2}kg\times1\times10^{-10}m}\)

= 3.5 × 10⁻²⁴ m/s.