Question

Suppose a proton and an electron were held together in a hydrogen atom by gravitational forces only. Find the formula for the energy levels of such an atom, the radius of its ground-state Bohr orbit, and its ionization energy in eV.

Answer 1

$mvr=\frac{nh}{2\pi}\to v=nh/(2\pi mr)$

$m\frac{v^2}{r}=G\frac{Mm}{r^2}\to v^2=GM/r$

So, we heve

$\frac{n^2h^2}{4\pi^2m^2r^2}=\frac{GM}{r}\to r_n=\frac{n^2h^2}{4\pi^2m^2GM}$ ($m$ - is the mass of the electron, $M$ - is the mass of the proton).

The radius of the electron ground-state Bohr orbit

$r_1=\frac{h^2}{4\pi^2m^2GM}=\frac{(6.62\cdot10^{-34})^2}{4\cdot3.14^2\cdot (9.1\cdot10^{-31})^2\cdot6.67\cdot10^{-11}\cdot 1.67\cdot10^{-27}}=1.2\cdot10^{29}\ (m)$

After performing similar calculations we get

$E_n=-\frac{2\pi^2G^2M^2m^3}{h^2n^2}=-\frac{2\cdot3.14^2\cdot(6.67\cdot10^{-11})^2\cdot(1.67\cdot10^{-27})^2\cdot(1.6\cdot10^{-31})^3}{(6.62\cdot10^{-34})^2\cdot1^2}=-4.2\cdot10^{-97}\ (J)=$

$=-2.6\cdot10^{-78}\ (eV)$

$E_i=E_{max}-E_{min}=0-(-2.6\cdot10^{-78})=2.6\cdot10^{-78}\ (eV)$