# How is most probable radius in quantum mechanical model equal to Bohr's radius in hydrogen atom

Most probable radius in 1s orbital for hydrogen electron is $$\pu{0.529E-10 m}$$ which is Bohr radius. But energy of electron in hydrogen atom is proportional to mean radial distance in atom. In this situation most probable radius is equal to average radius? What's going on?

• The mean and most probable values of a distribution, generally speaking, are not obliged to be related in any specific way. One may be less than the other, or greater, or equal to it. That's what is going on. On a side note, the energy is not proportional to the mean radius, but that's another story. – Ivan Neretin Jun 18 at 6:42

We find the radius at which the radial distribution function of the hydrogenic 1s orbital has a maximum value by solving $$\mathrm dP/\mathrm dr = 0.$$

If there are several maxima, then we choose the one corresponding to the greatest amplitude (the outermost one).

The radial distribution function is given by

$$P(r) = \frac{4Z^3r^2}{a_0^3}\exp\left(\frac{-2Zr}{a_0}\right).\tag{1}$$

It follows that

$$\frac{\mathrm dP}{\mathrm dr} = \frac{4Z^3}{a_0^3}\left(2r - \frac{2Zr^2}{a_0}\right)\exp\left(\frac{-2Zr}{a_0}\right).\tag{2}$$

This function is zero where the term in parentheses is zero, which is at

$$r_\mathrm{mp} = \frac{a_0}{Z}.\tag{3}$$

For hydrogen $$Z = 1,$$ so $$r_\mathrm{mp} = a_0.$$

For more details, see Atkins' Physical Chemistry.