Respuesta :
The question is incomplete. Here is the complete question.
An atom of lead has a radius of 154 pm and the average orbitalspeed of the electron in it is about 1.8x[tex]10^{8}[/tex] m/s. Calculate the least possible uncertainty in a measurement of the speed of an electron in an atom of lead. Write your answer as a percentage of the average speed, and round it to significant 2 digits.
Answer: v% = 0.21 m/s
Explanation: To calculate the uncertainty, use Heisenberg's Uncertainty Principle, which states that: ΔpΔx≥[tex]\frac{h}{4\pi }[/tex]
where h is Planck's constant and it is equal to 6.626.[tex]10^{-34}[/tex]m²kg/s.
Since p (momentum) is p = m.v:
mΔv.Δx ≥ [tex]\frac{h}{4\pi }[/tex]
Δv = [tex]\frac{h}{4\pi.x.m }[/tex]
Given that: r = x = 1.54.[tex]10^{-10}[/tex]m and mass of an electron is m=9.1.[tex]10^{-31}[/tex]kg
Δv = [tex]\frac{6.626.10^{-34} }{4.3.14.1.54.10^{-10}.9.1.10^{-31}}[/tex]
Δv = 0.0376.[tex]10^{7}[/tex]
As percentage of average speed:
Δv.[tex]\frac{1}{v}[/tex].100% = [tex]\frac{0.0376.10^{7} }{1.8.10^{8} }[/tex].10² = 0.021.10 = 0.21%
The least possible uncertainty in a speed of an electron is 0.21%.
