Question

(a) A proton is confined to the nucleus of an atom. Assume the nucleus has diameter 5.5 x 10-15 m and that this distance is the uncertainty in the proton's position. What is the minimum uncertainty in the momentum of the proton? Dpmin = kg-m/s

(b) An electron is confined in an atom. Assume the atom has diameter 1 x 10-10 m and that this distance is the uncertainty in the electron's position. What is the minimum uncertainty in the momentum of the electron?

Answer 1

`1.91738* 10^(-20)\ kgm/s`

`1.05456* 10^(-24)\ kgm/s`

Step-by-step explanation:

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

`\Delta x` = Uncertainty in the position

`\Delta p` = Uncertainty in the momentum

From the uncertainty principle

`\Delta x\Delta p=(h)/(2\pi)\\\Rightarrow \Delta p=(h)/(2\pi \Delta x)\\\Rightarrow \Delta p=(h)/(2\pi \Delta x)\\\Rightarrow \Delta p=(6.626* 10^(-34))/(2\pi* 5.5* 10^(-15))\\\Rightarrow \Delta p=1.91738* 10^(-20)\ kgm/s`

The minimum uncertainty in the momentum of the proton is
`1.91738* 10^(-20)\ kgm/s`

`\Delta x\Delta p=(h)/(2\pi)\\\Rightarrow \Delta p=(h)/(2\pi \Delta x)\\\Rightarrow \Delta p=(h)/(2\pi \Delta x)\\\Rightarrow \Delta p=(6.626* 10^(-34))/(2\pi* 1* 10^(-10))\\\Rightarrow \Delta p=1.05456* 10^(-24)\ kgm/s`

The minimum uncertainty in the momentum of the electron is
`1.05456* 10^(-24)\ kgm/s`