Radius of Bohr’s Stationary Orbit , Orbital Speed

Bohr’s Stationary Radius : An electron experiences the centripetal electrostatic force of attraction Fe exerted by the positively nucleus of charge Ze, (where Z = atomic number of the nucleus)

$ \displaystyle F_e = \frac{m v^2}{r} $

According to Coulomb’s Law :

$ \displaystyle \frac{1}{4\pi \epsilon_0}\frac{(Ze)(e)}{r^2} = \frac{Z e^2}{4\pi\epsilon_0 r^2}$

$ \displaystyle \frac{Z e^2}{4\pi\epsilon_0 r^2} = \frac{m v^2}{r} $

$ \displaystyle v^2 = \frac{Z e^2}{4\pi\epsilon_0 m r} $

Eliminating v by using the expression mvr = nh/2π  , n being an integer (n = 1, 2, 3, . . . ) we obtain,

$ \displaystyle (\frac{n h}{2\pi mr})^2 = \frac{Z e^2}{4\pi\epsilon_0 m r} $

$ \displaystyle \frac{n^2 h^2}{4\pi^2 m^2 r^2} = \frac{Z e^2}{4\pi\epsilon_0 m r} $

$ \displaystyle r = \frac{\epsilon_0 n^2 h^2}{\pi Z m e^2} $

Since , h, Z, m etc. are constants. Expressing r in terms of n and Z,

$ \displaystyle r = \frac{r_0 n^2}{Z} $ Where $latex \displaystyle r_0 = \frac{\epsilon_0 h^2}{\pi m e^2} $

is the radius of the orbit of an electron in its ground state

n = 1, of a hydrogen atom (Z = 1)

Numerically, r0 = 0.53 A0

Thus, $ \displaystyle r_0 = 0.53\frac{n^2}{Z}A^o $

Orbital speed:

$ \displaystyle mvr = \frac{nh}{2\pi} $

$ \displaystyle v = \frac{nh}{2\pi m r} $

Substituting the value of r we get

$ \displaystyle v = nh/(2\pi m)\frac{\epsilon_0 n^2h^2}{\pi m e^2 Z} $

$ \displaystyle v = \frac{Z e^2}{2\epsilon_0 n h} $

$ \displaystyle v = \frac{v_0 Z}{n} $

where v0 = e2/2ε0h is the speed of an electron in Bohr’s ground state (n = 1) orbit of hydrogen atom.

Also Read :

∗ Rutherford experiment & Observations
∗ Rutherford experiment : Conclusion & Limitations
∗ Bohr’s Atomic Theory
∗ Energy of the electron in nth orbit
∗ Origin of Spectra

← Back Page | Next Page → 

Leave a Reply