# Can the ionisation energy of hydrogen be calculated using Coloumb’s Law?

Using Coloumb’s Law: $$F=k\frac{qQ}{r^2}$$

and the potential energy function: $$U(r)=\int_\infty^r F dr$$

I get:

$$U(r)=k\frac{qQ}{r}$$

Where k is the Coloumb constant and q and Q are the values of negative and positive charge respectively.

Inputting the values of charge for one electron and one proton for a hydrogen atom, the value of k and the Bohr radius I get:

$$U=8.99\times10^9 \frac{-1.60\times10^{-19} \times1.60\times10^{-19}}{5.29\times10^{-11}}$$

Which works out as:

$$U=4.35\times10^{-18}J$$

Multiplying this by the Avogadro number, dividing by 1000 and rounding up gives:

$$U=2600 kJ mol^{-1}$$

This value is approximately double that of the actual ionisation energy of hydrogen and closer to that of helium but I can’t see the reason for it.

• Hello! Good try to come up with a new idea to calculate something. Unfortunately, Bohr did it like a $100$ years ago, combining Newton's second law with Coulomb's law, and also got a value of the ionization energy. You may want to read the Bohr's model... Commented Aug 24, 2023 at 17:07
• You didn't consider the energy electron already has. Commented Aug 24, 2023 at 18:04

Yes, the ionization energy of hydrogen can be calculated using Coulomb's law for the potential energy, but you must also take into account the kinetic energy. In Bohr's model, the electron in the ground state is spinning around the nucleus at around 1% of the speed of light. As you bring this electron to infinity (=ionization) it progressively loses this speed, since overall angular momentum $$mvr$$ is conserved.
If you do want to calculate the kinetic energy explicitly, you'll need to know the speed of the electron, which follows from Bohr's model that you'll have to look up elsewhere. I'll just give the result here: the speed of the electron is $$\alpha$$ times the speed of light, where $$\alpha$$ is the so-called fine-structure constant that is approximately 1/137. You can then state the kinetic energy in the ground state as : $$E = \frac12\cdot m\cdot v^2$$ Entering the electron mass, the speed of light and the fine structure constant gives $$2.18 \times 10^{-18} J$$, which is indeed half of what you calculated for the potential energy above.