Question

An atom of carbon has a radius of 67.0 pm and the average orbital speed of the electrons in it is about 1.3 x 10⁶ m/s.

Answer 1

Answer :

The least possible uncertainty in an electron's velocity is, $4.32\times 10^{5}m/s$

The percentage of the average speed is, 33 %

Explanation :

According to the Heisenberg's uncertainty principle,

$\Delta x\times \Delta p=\frac{h}{4\pi}$ ...........(1)

where,

$\Delta x$ = uncertainty in position

$\Delta p$ = uncertainty in momentum

h = Planck's constant

And as we know that the momentum is the product of mass and velocity of an object.

$p=m\times v$

or,

$\Delta p=m\times \Delta v$      .......(2)

Equating 1 and 2, we get:

$\Delta x\times m\times \Delta v=\frac{h}{4\pi}$

$\Delta v=\frac{h}{4\pi \Delta x\times m}$

Given:

m = mass of electron = $9.11\times 10^{-31}kg$

h = Planck's constant = $6.626\times 10^{-34}Js$

radius of atom = $67.0pm=67.0\times 10^{-12}m$     $(1pm=10^{-12}m)$

$\Delta x$ = diameter of atom = $2\times 67.0\times 10^{-12}m=134.0\times 10^{-12}m$

Now put all the given values in the above formula, we get:

$\Delta v=\frac{6.626\times 10^{-34}Js}{4\times 3.14\times (134.0\times 10^{-12}m)\times (9.11\times 10^{-31}kg)}$

$\Delta v=4.32\times 10^{5}m/s$

The minimum uncertainty in an electron's velocity is, $4.32\times 10^{5}m/s$

Now we have to calculate the percentage of the average speed.

Percentage of average speed = $\frac{\text{Uncertainty in speed}}{\text{Average speed}}\times 100$

Uncertainty in speed = $4.32\times 10^{5}m/s$

Average speed = $1.3\times 10^{6}m/s$

Percentage of average speed = $\frac{4.32\times 10^{5}m/s}{1.3\times 10^{6}m/s}\times 100$

Percentage of average speed = 33.2 % ≈ 33 %

Thus, the percentage of the average speed is, 33 %