# Bohri aatomimudeli matemaatilised alused.

Kui elektron liigub stabiilsel orbitaalil ümber aatomituuma, siis elektronile mõjuv tsentrifugaaljõud on võrdne elektrostaatilise jõuga tuuma ja elektroni vahel:

 m v2 / r   =   e2 / (4 p e0 r2)

kus
m .... elektroni mass
v .... elektroni kiirus
e .... elementaarlaeng (elektroni laeng)
e0 ... vaakumi dielektriline läbitavus

Aga Bohri mudelis on lubatud vaid need orbitaali raadiused, mille impulsimomedid on täisarvkordsed h/(2p)-st:

Bohri kvanttingimus:

 r m v   =   n h / (2p)

m ... elektroni mass
v ... elektroni kiirus
n ... kvantarv (n = 1, 2, 3, ...)
h ... Planck'i konstant

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 l   =   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 p   =   n h / (m v), which proves the above-mentioned 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:

 r   =   (h2 e0 / (m e2 p)) · n2

h .... Planck's constant
e0 ... permittivity of vacuum
m .... mass of the electron
e .... elementary charge
n .... principal quantum number (n = 1, 2, 3, ...)

Using the formulation E   =   Epot + Ekin   =   - e2 / (4 p e0 r) + (m / 2) v2, we get:

Energy of the hydrogen atom for the state of principal quantum number n:

 E   = - (m e4 / (8 e0 2 h2)) · 1 / n2

m .... mass of the electron
e .... elementary charge
e0 ... 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 so-called reduced mass m' in the formula above:

Reduced mass of the electron:

 m' = mN m / (mN + m)

m .... mass of the electron
mN ... mass of the nucleus

URL: http://www.walter-fendt.de/ph14e/bohrmath_ee.htm
© Walter Fendt, May 29, 1999
© Translation: Kaido Reivelt, 2007 (www.fyysika.ee)
Last modification: January 30, 2010