# Why is nucleus charge squared in Rydbergs formula?

According to coulumbs law, the energy changes linearly with charge. So why is the nucleus charge squared in this formula? Is there a way to derive it?

I was suspecting it might be because it pulls the electron closer too.

• Your qualitative understanding is correct. As to the rigorous derivation, wait until you learn the internal structure of an atom, starting from the Bohr model (which then should be forgotten, but that's another story). – Ivan Neretin Mar 28 at 19:48

Derivation comes from the formula for electrostatic potential energy.

The electrostatic potential between the point charges of an electron, and a proton would be represented by the formula

$$U_E(r) = k_e \frac{q_eQ_p}{r}$$

However, in the derivation that we are looking at a difference of energy is from an emission or absorption, such that the energy concerned is

$$E_i - E_f$$

The subtraction appears in the Rydberg formula from the distance in the form

$$\left( \frac {1} {n_1^2} - \frac {1} {n_2^2} \right)$$

However, the terms other than the radius are multiplied together and come outside the parentheses. Even though we are looking at only a single electron, the term for the charge in the nucleus is squared because of this. This can be seen in the Bohr derivation of the Rydberg formula.

$$E = h\nu = E_i - E_f = \frac {m_eq_e^2q_Z^2} {8h^2\epsilon_0} \left( \frac {1} {n_f^2} - \frac {1} {n_i^2} \right)$$

This shows the origin of the squared term for the nuclear charge which correlates with the Rydberg formula for a hydrogen-like element since all terms outside the parentheses other than charge are constant.

$$\frac {1} {\lambda_{vac}} = RZ^2\left( \frac {1} {n_1^2} - \frac {1} {n_2^2} \right)$$

Strictly speaking the nuclear charge $$Z$$ is not included in Rydberg's constant but multiplies this as $$Z^2$$ when used in the Balmer formula for a transition frequency $$\displaystyle \nu = Z^2R_\infty\left( \frac{1}{n_i^2}-\frac{1}{n_f^2} \right)$$ between levels $$i$$ and $$f$$. In a series of x-ray experiments Moseley, in 1914, was the first to confirm that the atomic number $$Z$$ was proportional to the square root of the transition frequency.

The $$Z^2$$ comes about in the Bohr model by considering the electrostatic force between the electron and nucleus, this is (in SI units) $$\displaystyle \frac{Ze^2}{(4\pi\epsilon_o)}\frac{1}{r^2}$$. This is the derivative of the electrostatic energy when charges are separated by a distance $$r$$. The nucleus has a charge $$Ze$$, the electron just $$1e$$.

This force is then equated with the centripetal acceleration giving $$\displaystyle \frac{Ze^2}{(4\pi\epsilon_o)}\frac{1}{r^2}=\frac{m_ev^2}{r}$$ where $$m_e$$ is the electron mass and $$v$$ its speed. The total energy is the sum of kinetic and potential energy and so $$\displaystyle E=\frac{1}{2}m_ev^2-\frac{Ze^2}{(4\pi\epsilon_o)}\frac{1}{r}$$. Finally, Bohr assumed that only certain orbits were allowed and the equation for the energy leads to the transition frequency formula given above with $$\displaystyle hcR_\infty=\left(\frac{e^2}{(4\pi\epsilon_o)}\right)^2\frac{m_e}{2\hbar^2}$$ and $$hc$$ converts from energy to wavenumbers.

In the proper, quantum, derivation the same Coulomb electrostatic energy is used in the Schroedinger equation as was used by Bohr and is $$\displaystyle -\frac{Ze^2}{(4\pi\epsilon_o)}\frac{1}{r}$$