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

Answer:

$1.91738\times 10^{-20}\ kgm/s$

$1.05456\times 10^{-24}\ kgm/s$

Explanation:

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

$\Delta x$ = Uncertainty in the position

$\Delta p$ = Uncertainty in the momentum

From the uncertainty principle

$\Delta x\Delta p=\dfrac{h}{2\pi}\\\Rightarrow \Delta p=\dfrac{h}{2\pi \Delta x}\\\Rightarrow \Delta p=\dfrac{h}{2\pi \Delta x}\\\Rightarrow \Delta p=\dfrac{6.626\times 10^{-34}}{2\pi\times 5.5\times 10^{-15}}\\\Rightarrow \Delta p=1.91738\times 10^{-20}\ kgm/s$

The minimum uncertainty in the momentum of the proton is $1.91738\times 10^{-20}\ kgm/s$

$\Delta x\Delta p=\dfrac{h}{2\pi}\\\Rightarrow \Delta p=\dfrac{h}{2\pi \Delta x}\\\Rightarrow \Delta p=\dfrac{h}{2\pi \Delta x}\\\Rightarrow \Delta p=\dfrac{6.626\times 10^{-34}}{2\pi\times 1\times 10^{-10}}\\\Rightarrow \Delta p=1.05456\times 10^{-24}\ kgm/s$

The minimum uncertainty in the momentum of the electron is $1.05456\times 10^{-24}\ kgm/s$