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 * 10^(-24) \mathrm{m} / \mathrm{s}`

Step-by-step 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 * \Delta p=(h)/(4 \pi)`

Here momentum is the product of mass and velocity.

So,
`\Delta x *(m * \Delta v)=(h)/(4 \pi)`

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

position delta x =
`1 * 10^(10)`

Planck's constant h =
`6.626 * 10^(-34)`
`k g \cdot m^(2) / s`

`\Delta v=(h)/(4 \pi * \Delta x * m)=(6.626 * 10^(-34))/(4 * 3.14 * 1 * 10^(10) * 0.150)`

`\Delta v=(6.626 * 10^(-34+10))/(4 * 3.14 * 0.150)=(6.626 * 10^(-24))/(1.884)`

`\Delta v=3.51 * 10^(-24) \mathrm{m} / \mathrm{s}`

So the uncertainty in velocity is
`3.51 * 10^(-24) \mathrm{m} / \mathrm{s}`.