Respuesta :
The question is incomplete, here is the complete question:
An atom of helium has a radius of 31. pm and the average orbital speed of the electrons in it is about [tex]4.4\times 10^6m/s[/tex]. Calculate the least possible uncertainty in a measurement of the speed of an electron in an atom of helium. Write your answer as a percentage of the average speed, and round it to 2 significant digits
Answer: The percentage of average speed is 41 %
Explanation:
We are given:
Radius of helium atom = 31 pm = [tex]31\times 10^{-12}m[/tex] (Conversion factor: [tex]1m=10^{12}pm[/tex] )
So, diameter of helium atom = [tex](2\times r)=(2\times 31\times 10^{-12})=64\times 10^{-12}m[/tex]
The diameter of the atom will be equal to the uncertainty in position.
The equation representing Heisenberg's uncertainty principle follows:
[tex]\Delta x.\Delta p=\frac{h}{2\pi}[/tex]
where,
[tex]\Delta x[/tex] = uncertainty in position = d = [tex]64\times 10^{-12}m[/tex]
[tex]\Delta p[/tex] = uncertainty in momentum = [tex]m\Delta v[/tex]
m = mass of electron = [tex]9.1095\times 10^{-31}kg[/tex]
h = Planck's constant = [tex]6.627\times 10^{-34}kgm^2/s^2[/tex]
Putting values in above equation, we get:
[tex]64\times 10^{-12}m\times 9.1095\times 10^{-31}kg\times \Delta v=\frac{6.627\times 10^{-34}kgm^2/s}{2\times 3.14}\\\\\Delta v=\frac{6.627\times 10^{-34}kgm^2/s^2}{2\times 3.14\times 64\times 10^{-12}m\times 9.1095\times 10^{-31}kg}=1.81\times 10^6m/s[/tex]
To calculate the percentage of average speed, we use the equation:
[tex]\text{Percentage of the average speed}=\frac{\text{Uncertainty in velocity}}{\text{Average orbital speed}}\times 100[/tex]
We are given:
Average orbital speed = [tex]4.4\times 10^6m/s[/tex]
Putting values in above equation, we get:
[tex]\text{Percentage of the average speed}=\frac{1.81\times 10^6m/s}{4.4\times 10^6m/s}\times 100\\\\\text{Percentage of the average speed}=41.\%[/tex]
Hence, the percentage of average speed is 41 %
