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Question

The electrostatic force of attraction between a proton and an electron in a hydrogen atom 
corresponds to which component of energy possessed by the electron?


Solution

You can calculate the Coloumb force between any two charged bodies using Coulomb’s Law:

F = k (q1*q2/r^2)

Where k is Coloumb’s constant:

8.99*10^9Nm^2C^-2

The electron carries the elementary chargeq = -e.

The proton carries the same charge but positiveq = e

The value of e is empirically measured to be: 1.602*10^-19 C

Now, the r value is a little tricky. r is meant to be the distance between charges. Unfortunately, thanks to quantum physics, the electron could be any distance from the proton. It only has a certain probability of being a given distance.

Luckily, physicists account for this by giving a value called the Bohr radius. This is the average or most likely distance a single electron would be from the centre of the single proton in a hydrogen atom. This value is defined as: 5.291772*10^-11m

So we have all our values! Let's plug ’em in and see what we get.

F = k (q1*q2/r^2)

F = (8.99*10^9Nm^2C^-2)(((-1.602*10^-19 C)(1.602*10^-19 C))/(5.291772*10^-11m)^2)

F = (8.99*10^9)(-9.141*10^-18)

F = -8.22*10^-8N

Note that the value is negative because the proton and electron are attracted to each other. As well, the value would vary depending on where the electron was at any given moment and so this is only the most probable value.

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