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.

Calculate the least possible uncertainty in a measurement of the speed of an electron in an atom of carbon. Write your answer as a percentage of the average speed, and round it to significant digits.

Answer 1

The least possible uncertainty in an electron's velocity is,
`4.32* 10^(5)m/s`

The percentage of the average speed is, 33 %

Explanation :

According to the Heisenberg's uncertainty principle,

`\Delta x* \Delta p=(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* v`

or,

`\Delta p=m* \Delta v` .......(2)

Equating 1 and 2, we get:

`\Delta x* m* \Delta v=(h)/(4\pi)`

`\Delta v=(h)/(4\pi \Delta x* m)`

Given:

m = mass of electron =
`9.11* 10^(-31)kg`

h = Planck's constant =
`6.626* 10^(-34)Js`

radius of atom =
`67.0pm=67.0* 10^(-12)m`
`(1pm=10^(-12)m)`

`\Delta x` = diameter of atom =
`2* 67.0* 10^(-12)m=134.0* 10^(-12)m`

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

`\Delta v=(6.626* 10^(-34)Js)/(4* 3.14* (134.0* 10^(-12)m)* (9.11* 10^(-31)kg))`

`\Delta v=4.32* 10^(5)m/s`

The minimum uncertainty in an electron's velocity is,
`4.32* 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}}* 100`

Uncertainty in speed =
`4.32* 10^(5)m/s`

Average speed =
`1.3* 10^(6)m/s`

Percentage of average speed =
`(4.32* 10^(5)m/s)/(1.3* 10^(6)m/s)* 100`

Percentage of average speed = 33.2 % ≈ 33 %

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