Respuesta :
Answer :
The least possible uncertainty in an electron's velocity is, [tex]4.32\times 10^{5}m/s[/tex]
The percentage of the average speed is, 33 %
Explanation :
According to the Heisenberg's uncertainty principle,
[tex]\Delta x\times \Delta p=\frac{h}{4\pi}[/tex] ...........(1)
where,
[tex]\Delta x[/tex] = uncertainty in position
[tex]\Delta p[/tex] = uncertainty in momentum
h = Planck's constant
And as we know that the momentum is the product of mass and velocity of an object.
[tex]p=m\times v[/tex]
or,
[tex]\Delta p=m\times \Delta v[/tex] .......(2)
Equating 1 and 2, we get:
[tex]\Delta x\times m\times \Delta v=\frac{h}{4\pi}[/tex]
[tex]\Delta v=\frac{h}{4\pi \Delta x\times m}[/tex]
Given:
m = mass of electron = [tex]9.11\times 10^{-31}kg[/tex]
h = Planck's constant = [tex]6.626\times 10^{-34}Js[/tex]
radius of atom = [tex]67.0pm=67.0\times 10^{-12}m[/tex] [tex](1pm=10^{-12}m)[/tex]
[tex]\Delta x[/tex] = diameter of atom = [tex]2\times 67.0\times 10^{-12}m=134.0\times 10^{-12}m[/tex]
Now put all the given values in the above formula, we get:
[tex]\Delta v=\frac{6.626\times 10^{-34}Js}{4\times 3.14\times (134.0\times 10^{-12}m)\times (9.11\times 10^{-31}kg)}[/tex]
[tex]\Delta v=4.32\times 10^{5}m/s[/tex]
The minimum uncertainty in an electron's velocity is, [tex]4.32\times 10^{5}m/s[/tex]
Now we have to calculate the percentage of the average speed.
Percentage of average speed = [tex]\frac{\text{Uncertainty in speed}}{\text{Average speed}}\times 100[/tex]
Uncertainty in speed = [tex]4.32\times 10^{5}m/s[/tex]
Average speed = [tex]1.3\times 10^{6}m/s[/tex]
Percentage of average speed = [tex]\frac{4.32\times 10^{5}m/s}{1.3\times 10^{6}m/s}\times 100[/tex]
Percentage of average speed = 33.2 % ≈ 33 %
Thus, the percentage of the average speed is, 33 %
