Question

Suppose that you know the position of a 100-gram pebble to within the width of an atomic nucleus ( δx=10−15δx=10−15 meters). what is the minimum uncertainty in the momentum of the pebble?

Express your answer in kilogram meters per second to one significant figure.
1 * 10-19 or 5 *10-20 are not correct answers

Answer 1

The minimum uncertainty in the momentum of the pebble is
`5.27 x 10^{-20` kg m/s.

Step-by-step explanation:

The Heisenberg Uncertainty Principle states that there is a trade-off between the precision of our knowledge of an object's position and its momentum. The uncertainty in momentum (Δp) is related to the uncertainty in position (Δx) by the equation ΔxΔp ≥ ħ/2, where ħ is the reduced Planck's constant.

In this case, the position uncertainty (Δx) is given as
`10^{-15` meters. To find the minimum uncertainty in momentum, we can rearrange the equation:

Δp ≥ ħ/2Δx

Plugging in the values:

Δp ≥
`(1.055 * 10^(-34) kg m²/s) / (2 * 10^(-15) m)`

Calculating this:

Δp ≥ 5.27 x
`10^{-20` kg m/s

Therefore, the minimum uncertainty in momentum of the pebble is 5.27 x
`10^{-20` kg m/s.

Answer 2

The minimum uncertainty in the momentum of the pebble is 3.313 × 10^-19 kg m/s.

Step-by-step explanation:

The uncertainty in momentum, Δp, can be calculated using the equation ΔxΔp ≥ ħ/2, where Δx is the uncertainty in position and ħ is the reduced Planck's constant. Given that the uncertainty in position, Δx, is 10^-15 meters and the mass of the pebble, m, is 100 grams, we can calculate the uncertainty in momentum as Δp = ħ/(2Δx).

Converting the mass of the pebble to kilograms, we get m = 0.1 kg. Plugging in the values, we have Δp = (6.626 × 10^-34 kg m²/s)/(2 × 10^-15 m) = 3.313 × 10^-19 kg m/s.

The minimum uncertainty in the momentum of the pebble is 3.313 × 10^-19 kg m/s.