The negatively charged electron of the hydrogen atom is forced to a circular motion by the attractive electrostatic force of the positively charged atomic nucleus. Thus, the electrostatic force is the centripetal force.
m v^{2} r 
=  e^{2} 4 π ε_{0} r^{2} 
m .... mass of the electron
v .... velocity of the electron
r .... orbital radius
e .... elementary charge
ε_{0} ... permittivity of vacuum
However, only those orbital radii are allowed, for which the angular momentum is an integer multiple of h / (2 π).
Bohr's quantum condition:

r ... orbital radius
m ... mass of the electron
v ... velocity of the electron
n ... principal quantum number (n = 1, 2, 3, ...)
h ... Planck's constant
Bohr's quantum condition sounds plausible, if one takes the idea of de Broglie waves (waves of matter) as a starting point: The electron corresponds to a wave of wavelength λ = h / (m v). For the existence of a standing wave around the nucleus it is necessary that the circumference of the orbit is an integer multiple of the wavelength. Thus we have 2 r π = n h / (m v), which proves the abovementioned quantum condition.
By solving the second equation for v and inserting the result into the first equation, we obtain the following result for the allowed radii:
Orbital radius for the state of principal quantum number n:

h .... Planck's constant
ε_{0} ... permittivity of vacuum
m .... mass of the electron
e .... elementary charge
n .... principal quantum number (n = 1, 2, 3, ...)
Using the formulation E = E_{pot} + E_{kin} = − e^{2} / (4 π ε_{0} r) + (m / 2) v^{2}, we get:
Energy of the hydrogen atom for the state of principal quantum number n:

m .... mass of the electron
e .... elementary charge
ε_{0} ... permittivity of vacuum
h .... Planck's constant
n .... principal quantum number (n = 1, 2, 3, ...)
Strictly speaking, it is necessary to make a small correction to this formula. The mass of the nucleus is certainly much bigger than that of the electron, but not infinite. Thus, electron and atomic nucleus revolve about their common center of gravity, which is not exactly identical with the center of the atom. If we take this into consideration, we have to replace the mass of the electron (m) by the socalled reduced mass m' in the formula above:
Reduced mass of the electron:

m .... mass of the electron
m_{N} ... mass of the nucleus
URL: http://www.walterfendt.de/ph14e/bohrmath_e.htm
© Walter Fendt, May 29, 1999
Last modification: February 4, 2010