Question

What is the instantaneous velocity of an electron orbiting a lithium atom at a radius of `90` picometres?

Answer 1

The key to this one is combining the expression for the centripetal force, `F=(mv^2)/r`, with Coulomb's law, `F=(kq_1q_1)/r^2`. Stir in with some constants, and see the full calculation below. The solution is `v=8.6xx10^12` `ms^-1`.

Explanation:

We can set these two expressions equal to each other, because it is the electrostatic (Coulomb's law) attraction that provides the centripetal force for the circular motion.

(side note: electrons don't actually orbit like planets, from a quantum perspective, but this question takes a classical approach)

`(mv^2)/r=(kq_1q_1)/r^2`

We want to rearrange this to make the speed, `v`, the subject.

Multiply both sides by `r` and cancel:

`mv^2=(kq_1q_1 cancel(r))/r^cancel(2)=(kq_1q_1)/r`

Divide both sides by `m`:

`v^2=(kq_1q_1)/(mr)`

Take the square root of both sides:

`v=sqrt((kq_1q_1)/(mr))`

OK, we're all ready now for the calculation. We have been told that `r=90` picometres = `90xx10^-12=9xx10^-11` `m`

The Coulomb's law constant, `k=9.0xx10^9` `Nm^2C^-2`

The charge on an electron is `-1.6xx10^-19` `C`. A lithium nucleus has 3 protons, so its charge is `3xx1.6xx10^-19=4.8xx10^-19` `C`.

The opposite signs tell us that this is an attractive force, but we can ignore the sign in the calculation to find the magnitude of the speed.

The mass of an electron is `9.1xx10^-31` `kg`.

Plugging in all these values, we have:

`v=sqrt((9xx10^9xx1.6xx10^-19xx4.8xx10^-19)/(9.1xx10^-31xx9.0xx10^-11))`

My calculator tells me that all that yields `v=8.6xx10^12` `ms^-1`.

Answer 2

`approx2.9xx10^5ms^-1`

Explanation:

We can use Classical mechanics to find the desired value of speed `|vec v|` by equating Coulomb's force of attraction between the electron and the lithium nucleus to the centripetal force experienced by the electron due to its circular motion.

`(m_e|vecv|^2)/r_e=(kq_eq_n)/r_e^2`
Where `m_e` is mass of an electron`=9.1xx10^-31kg`,
`r_e` radial distance between the electron and lithium nucleus,
`q_e` charge of an electron `=-1.6xx10^-19C`,
`q_n` charge of lithium nucleus,
`k` is Coulomb's constant `=9.0xx10^9` `Nm^2C^-2`,

Solving for `|vecv|`

`=>|vecv|^2=(kq_eq_n)/(m_er_e)`, and

`|vecv|=sqrt((kq_eq_n)/(m_er_e))` .....(1)

Given `r_e=90` pm `=90xx10^-12=9.0xx10^-11m`, a lithium nucleus has `3` protons, so has charge `q_n=3xx1.6xx10^-19` `=4.8xx10^-19C`.

Plugging all values in (1), we obtain

`|vecv|=sqrt(((9xx10^9)xx(1.6xx10^-19)xx(4.8xx10^-19))/((9.1xx10^-31)xx(9.0xx10^-11)))`
or `|vecv|approx2.9xx10^5ms^-1`.

we have ignored the sign of electron's charge in the calculation as we know that it only tells the direction of Coulomb's force, attractive here.