Answer :
Answer:
Percentage of uncertainty in average speed of an electron is 0.1756%.
Explanation:
Using Heisenberg uncertainty principle:
[tex]\Deltax\times \Delta p\geq \frac{h}{4\pi }[/tex]
[tex]\Delta p=m\times \Delta v[/tex]
[tex]\Deltax\times m\times \Delta v\geq \frac{h}{4\pi }[/tex]
Δx = Uncertainty in position
Δp = Uncertainty in momentum
Δv = Uncertainty in average speed
h = Planck's constant = [tex]6.626\times 10^{-34} kg m^2/s[/tex]
m = mass of electron = [tex]9.1\times 10^{-31} kg[/tex]
We have
Δx = 2 × 165 pm = 330 pm = [tex]3.3\times 10^{-10} m [/tex]
[tex]1 pm = 10^{-12} m[/tex]
Average orbital speed of electron = v = [tex]=1.0\times 10^8 m/s[/tex]
[tex]3.3\times 10^{-10} 9.1\times 10^{-31} kg \times \Delta v\geq \frac{6.626\times 10^{-34} kg m^2/s}{4\pi }[/tex]
[tex]\Delta v\geq \frac{6.626\times 10^{-34} kg m^2/s}{4\pi \times 3.3\times 10^{-10}\times 9.1\times 10^{-31} kg}[/tex]
[tex]\Delta v\geq 1.756\times 10^5 m/s[/tex]
Percentage of uncertainty in average speed:
[tex]=\frac{\Delta v}{v}\times 100[/tex]
[tex]=\frac{1.756\times 10^5 m/s}{1.0\times 10^8 m/s}\times 100=0.1756\%[/tex]