Question

An object is moving along a straight line, and the uncertainty in its position is 1.90 m.

Required:
Find the minimum uncertainty in the momentum of the object. Find the minimum uncertainty in the object's velocity, assuming that the object is (b) a golf ball (mass=0.045 kg) and (c) an electron.

Answer 1

The uncertainty principle states that there is a limit to how precisely we can know both the position and momentum of an object. The minimum uncertainty in momentum can be calculated using the uncertainty in position and the reduced Planck's constant.

Step-by-step explanation:

The uncertainty principle states that there is a limit to how precisely we can know both the position and momentum of an object. The uncertainty in momentum, Δp, is given by Δp ≥ ħ/2Δx, where Δx is the uncertainty in position and ħ is the reduced Planck's constant.

For a golf ball with a mass of 0.045 kg, the minimum uncertainty in momentum can be calculated using the uncertainty in position, which is 1.90 m. So, Δp ≥ (6.63 × 10^-34 kg m^2/s) / (2 × 1.90 m).

For an electron, the minimum uncertainty in momentum can be calculated in the same way, but the mass of an electron is 9.11 × 10^-31 kg. So, Δp ≥ (6.63 × 10^-34 kg m^2/s) / (2 × 1.90 m).

Answer 2

`2.78* 10^(-35)\ \text{kg m/s}`

`6.178* 10^(-34)\ \text{m/s}`

`0.31* 10^(-4)\ \text{m/s}`

Step-by-step explanation:

`\Delta x` = Uncertainty in position = 1.9 m

`\Delta p` = Uncertainty in momentum

h = Planck's constant =
`6.626* 10^(-34)\ \text{Js}`

m = Mass of object

From Heisenberg's uncertainty principle we know

`\Delta x\Delta p\geq (h)/(4\pi)\\\Rightarrow \Delta p\geq (h)/(4\pi\Delta x)\\\Rightarrow \Delta p\geq (6.626* 10^(-34))/(4\pi* 1.9)\\\Rightarrow \Delta p\geq 2.78* 10^(-35)\ \text{kg m/s}`

The minimum uncertainty in the momentum of the object is
`2.78* 10^(-35)\ \text{kg m/s}`

Golf ball minimum uncertainty in the momentum of the object

`m=0.045\ \text{kg}`

Uncertainty in velocity is given by

`\Delta p\geq m\Delta v\geq 2.78* 10^(-35)\\\Rightarrow \Delta v\geq (2.78* 10^(-35))/(m)\\\Rightarrow \Delta v\geq (2.78* 10^(-35))/(0.045)\\\Rightarrow \Delta v\geq 6.178* 10^(-34)\ \text{m/s}`

The minimum uncertainty in the object's velocity is
`6.178* 10^(-34)\ \text{m/s}`

Electron

`m=9.11* 10^(-31)\ \text{kg}`

`\Delta v\geq (\Delta p)/(m)\\\Rightarrow \Delta v\geq (2.78* 10^(-35))/(9.11* 10^(-31))\\\Rightarrow \Delta v\geq 0.31* 10^(-4)\ \text{m/s}`

The minimum uncertainty in the object's velocity is
`0.31* 10^(-4)\ \text{m/s}`.