What atomic number of an element "X" would have to become so that the 4th orbit around X would fit inside the 1st Bohr orbit of H-atom?

Aug 17, 2017

Well, the radial distance from the nucleus of a hydrogen-like atom is given by

$r = {n}^{2} \left({a}_{0} / Z\right)$.

Interestingly enough, one often sees the substitution

$\sigma = \frac{Z r}{{a}_{0}}$

in hydrogen atom wave functions. I derive this result below. Upon solving this you should get $Z = 16$.

We are asking, what would the atomic number $Z$ have to be such that the distance to the nucleus from $n = 4$ in a hydrogen-like atom (of the appropriate positive charge) is ${a}_{0}$, the Bohr radius. This is more of a numerical exercise than anything realistic...

For a given Bohr orbit, the wavelength is given by

$2 \pi r = n {\lambda}_{e}$ $\text{ } \boldsymbol{\left(1\right)}$

where $r$ is the radial distance from the nucleus, $n$ is the principal quantum number, and ${\lambda}_{e}$ is the electron wavelength.

Since electrons have rest mass ${m}_{e}$, we can invoke the de Broglie relation:

${\lambda}_{e} = \frac{h}{{m}_{e} v}$ $\text{ } \boldsymbol{\left(2\right)}$

where $v$ is its velocity and $h = 6.626 \times {10}^{- 34} \text{J"cdot"s}$ is Planck's constant.

We can thus rewrite $\left(1\right)$ as:

$2 \pi r = n \left(\frac{h}{{m}_{e} v}\right)$

=> vr = (nℏ)/m_e, $\text{ } \boldsymbol{\left(3\right)}$

where ℏ = h//2pi is the reduced Planck's constant.

Next, the coulomb force for a hydrogen-like atom is given by the derivative of the potential $V = - \frac{Z {e}^{2}}{4 \pi {\epsilon}_{0} r}$ with respect to the radial distance.

$- \frac{\partial V}{\partial r} = - \frac{\partial}{\partial r} \left[- \frac{Z {e}^{2}}{4 \pi {\epsilon}_{0} r}\right]$

$= - \frac{Z {e}^{2}}{4 \pi {\epsilon}_{0} {r}^{2}}$ $\text{ } \boldsymbol{\left(4\right)}$,

where ${\epsilon}_{0} = 8.854 \times {10}^{- 12} \text{F/m}$ is the vacuum permittivity and $e = 1.602 \times {10}^{- 19} \text{C}$ is the elementary charge.

From uniform circular motion in physics, the centripetal force (in the Bohr picture) brought about by the coulomb force (with inwards, or attraction, being negative) is given by

${F}_{c} = - \frac{{m}_{e} {v}^{2}}{r}$. $\text{ } \boldsymbol{\left(5\right)}$

Equating $\left(4\right)$ and $\left(5\right)$ (in the Bohr picture):

$\frac{{m}_{e} {v}^{2}}{r} = \frac{Z {e}^{2}}{4 \pi {\epsilon}_{0} {r}^{2}}$

$\implies \frac{Z {e}^{2}}{4 \pi {\epsilon}_{0} {m}_{e}} = {v}^{2} r$ $\text{ } \boldsymbol{\left(6\right)}$

Comparing ${\left(3\right)}^{2}$ above with $\left(6\right)$:

v^2r^2 = ((nℏ)/m_e)^2 = (Ze^2)/(4pi epsilon_0m_e) cdot r

Now if we solve this for $r$, we get...

r = ((nℏ)/m_e)^2 cdot (4pi epsilon_0m_e)/(Ze^2)

= n^2 ((4pi epsilon_0 ℏ^2)/(m_e e^2)) cdot 1/Z

By comparison with Wikipedia, we use the definition of the Bohr radius, ${a}_{0}$, to get:

$\textcolor{b l u e}{\overline{\underline{| \stackrel{\text{ ")(" "r = n^2 (a_0/Z)" }}{|}}}}$

So for a hydrogen-like atom, if we want $r = {a}_{0}$, we must have:

$Z = \frac{{n}^{2} \cdot {a}_{0}}{r}$

$= {n}^{2} \cdot {\cancel{{a}_{0} / {a}_{0}}}^{1}$

For the fourth Bohr orbit, we have $n = 4$, so $\textcolor{b l u e}{Z = 16}$. Thus we are looking at the sulfur cation, ${\text{S}}^{15 +}$, which is ridiculous to accomplish in real life!