Question

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

Answer 1

The uncertainty in velocity is $3.51 \times 10^{-24} \mathrm{m} / \mathrm{s}$

Explanation:

As per Heisenberg's uncertainity principle, the position and momentum of any object cannot be measured simultaneously. So product of uncertainty in position and momentum will be equal to modified plank's constant.

          $\Delta x \times \Delta p=\frac{h}{4 \pi}$

Here momentum is the product of mass and velocity.

So, $\Delta x \times(m \times \Delta v)=\frac{h}{4 \pi}$

Here, given mass (m) = 150 g = 0.150 kg

position delta x = $1 \times 10^{10}$

Planck's constant h = $6.626 \times 10^{-34}$ $k g \cdot m^{2} / s$

 $\Delta v=\frac{h}{4 \pi \times \Delta x \times m}=\frac{6.626 \times 10^{-34}}{4 \times 3.14 \times 1 \times 10^{10} \times 0.150}$

 $\Delta v=\frac{6.626 \times 10^{-34+10}}{4 \times 3.14 \times 0.150}=\frac{6.626 \times 10^{-24}}{1.884}$

 $\Delta v=3.51 \times 10^{-24} \mathrm{m} / \mathrm{s}$

So the uncertainty in velocity is $3.51 \times 10^{-24} \mathrm{m} / \mathrm{s}$.