Sommerfield atomic model

The Sommerfield atomic model:

Assumption:

To explain the splitting of \(H_\alpha,H_\beta,H_\gamma\) lines in the spectrum of H- atom, sommerfield proposed an atom model. The assumption of this atomic model are:-

1. The path of electron around the nucleus of an atom is general eliptical. The circular orbit of Bohr's theory are special case of elliptical orbit.

2. While revolving in elliptical orbit there is variation of speed of electron due to which there is relativistic variation in the most of electron. Here this model is also known as relativistic sommerfield atomic model.

Consider (r,\(\phi\)) be the instantaneous position of electron with respect to nucleus being at one of the foci of ellipse. The quantization condition associated with two co-ordinate r and \(\phi\) are.

$$\oint p_\phi d_\phi= n_\phi h\dotsm(1)$$

and, $$\oint p_r d_r=n_r h\dotsm(2)$$

where, \(p_\phi\)= angular momentum associated with chage in \(\phi\)=\(mr^2\frac{d\phi}{dt}\)

\(n_\phi\)= angular or Azumuthal no

h= Planck's constant = \(6.64\cdot 10^{-34}\)

\(P_r\)= radial momentum \(m\frac{dr}{dt}\)

\(n_r\)= radial quantum number

The principle quantum number is related with \(n_\phi\) and \(n_r\) is \(n_\phi+n_r=n\dotsm(3)\)

Evalution of \(\oint p_\phi d_\phi=n_\phi h\)

$$\oint p_\phi d_\phi=n_\phi h$$

$$or, p_\phi \oint d_\phi=n_\phi h $$ [ \(\therefore\) \(p_\phi\) is constant, being angular moment in central force ]

$$or, 2\phi\cdot p_\phi= n_\phi h$$

$$ p_\phi= \frac{n_\phi h}{2\pi}\dotsm(4)$$

Evalution of \(\oint p_r dr= n_r h\)

$$p_r d_r= m.\frac{dr}{dt}\cdot dt$$

$$=m\frac{dr}{d\phi}\cdot \frac{d\phi}{dt}\cdot\frac{dr}{d\phi}\cdot d\phi$$

$$=m\biggl(\frac{dr}{d\phi}\biggr)^2\cdot\biggl(\frac{d\phi}{dt}\biggl)\cdot d\phi$$

$$=m\biggl(\frac{dr}{d\phi}\biggr)^2\cdot\biggl(\frac{p_\phi}{mr^2}\biggl)$$

$$ =\biggl(\frac 1r \frac{dr}{d\phi}\biggr)^2\cdot p_\phi d_\phi \dotsm(5)$$

We have from polar equation of ellipse,

$$\frac 1r= \frac{1+\epsilon\cos\phi}{a(1-\epsilon^2)}\dotsm(6)$$

where, a= semi-major axis

b= ecentricity

Taking log on both side

$$or, -logr=log(1+\epsilon cos\phi)-log[a(1-\epsilon^2)]$$

Differentiating with respect to \(\phi\), we get,

$$\frac{-d(log r)}{dr}\cdot \frac{dr}{d\phi}=\frac{d log(1+\epsilon cos\phi}{d(1+\epsilon cos\phi)}\cdot \frac{d(1+\epsilon cos\phi)}{d\phi}-0$$

$$\frac{-1}{r}\cdot\frac{dr}{d\phi}= \frac{1}{(1+\epsilon cos\phi)}\cdot (0-\epsilon sin\phi)$$

$$ \frac 1r \frac{dr}{d\phi}= \frac{\epsilon sin\phi}{(1+\epsilon cos\phi)}\dotsm(7)$$

From (5) and (7),

$$P_r dr= \biggl(\frac{\epsilon sin\phi}{1+\epsilon cos\phi}\biggr)^2 p_\phi d\phi$$

$$\oint p_r \cdot dr= \int_0^{2\pi}\biggl(\frac{\epsilon sin\phi}{1+\epsilon cos\phi}\biggr)^2 p_\phi d\phi\dotsm(8)$$

let, $$I= \int_0^{2\pi} \frac{{\epsilon}^2 sin^2\phi}{(1+\epsilon cos\phi)^2} d\phi \dotsm(9)$$

We have, \(d[uv]= udv+vdu\)

\( udv= d[uv]-vdu\)

\(\int u.dv= [uv]-\int vdu\dotsm(10)\)

put, \(\epsilon sin \phi=u\) , \(c cos\phi\cdot d\phi= du\) and \(dv= \frac{\epsilon sin \phi d\phi}{(1+ \epsilon cos \phi)^2}\) \( implies\;\; v=\frac{1}{1+\epsilon cos \phi}\)

Substituting above equation in (10)

$$I= \int_0^{2\pi} \epsilon sin\phi\cdot\frac{ \epsilon sin\phi. d\phi}{(1+ cos\phi)^2}$$

$$=\biggl[\frac{\epsilon sin \phi}{1+\epsilon cos \phi}\bigg]_0^{2\pi}- \int_0^{2\pi}\frac{\epsilon cos\phi d\phi}{1+ \epsilon cos\phi}$$

$$=-\int_0^{2\pi} \frac{\epsilon cos\phi}{1+cos \phi} \cdot d\phi$$

$$= \int_0^{2\pi}\biggl[ \frac{-(1+\epsilon cos\phi)}{1+\epsilon cos\phi}+ \frac{1}{1+\epsilon cos\phi}\biggr] d\phi$$

$$=\int_0^{2\pi}\frac{1}{1+\epsilon cos\phi}\cdot d\phi- \int_0^{2\pi} 1d\phi\dotsm(11)$$

Using the value of standerd integral

$$i.e\;\; \int_0^{2\pi}\frac{d\phi}{1+\epsilon cos\phi}= \frac{2\pi}{\sqrt{1-\epsilon^2}}$$

$$ I= \frac{2\pi}{\sqrt{1-\epsilon^2}}- [\phi]_0^{2\pi}$$

$$I= \frac{2\pi}{\sqrt{1-\epsilon^2}}-2\pi\dotsm(12)$$

Substituting equation (12) and (9) in (8)

$$\oint p_r dr= p_\phi \biggl( \frac{2\pi}{(1-\epsilon^2)^\frac{1}{2}}-2\pi\biggr)=n_rh$$

$$or, \oint p_r dr= \frac{n_\phi h}{2\pi}\cdot 2\pi\biggl(\frac{1}{(1-\epsilon^2)^{\frac 12}}-1\biggr)=n_rh$$

$$or, \frac{n_\phi}{(1-\epsilon^2)^{\frac 12}}= n_\phi+n_r$$

$$or,\frac{n_\phi}{(1-\epsilon^2)^{\frac 12}}= n\;\; [from eq^n (3)]$$

$$or,\frac{n_phi}{n}= (1-\epsilon)^{\frac 12}$$

Squaring both side,

$$\frac{n_\phi^2}{n^2}= (1-\epsilon^2)=\frac{b^2}{a^2}\;\;\;[ b^2= a^2(1-\epsilon^2]$$

$$\therefore\;\;\; \frac{n_\phi}{n}=\frac ba\dotsm(13)$$

Equation (13) determines the shape and number of allow electronic orbit.

Case(1):

When \(n_\phi = 0\Rightarrow b=0\), the path of electronic orbit stright line passing through nucleus. But electron doesn't pass through nucleus \((n_\phi=0)\).

Case(2):

Since b and a are always positive \(n_\phi\) can not be negative.

Case(3):

Since \(b\leq a \Rightarrow n_\phi \leq n\)

When \(b=a\Rightarrow n_\phi=n\) The path of electronic orbit is circular.

When \(b\leq a\Rightarrow n_\phi\leq n \) The path of electronic orbit is eliptical.

Case(4) :

\(b\ngtr a\Rightarrow n_\phi\) can not exceed n

For example (1) When \(n=1\Rightarrow n_\phi =1\). The orbit is circular \(n_r=0\). This is the case of Bhor's theory.

(2) When \(n=2\Rightarrow n_\phi= 2\) Two orbit possible.

\(n=2, n_\phi=2 \Rightarrow a=b\) Orbit is circular or

\(n=2, n_\phi=1\Rightarrow b<a \)Orbit is elliptical

\(\Rightarrow b= \frac a2\)

(3) When \(n=3,n_\phi=3,2,1\) Theree possible orbit

\( n=3, n_\phi=3\Rightarrow b=a \) Orbit circular

\( n=3, n_\phi=2\Rightarrow b=\frac 23 a \) Orbit is eliptical

\( n=3, n_\phi=1\Rightarrow b=\frac 13 a \)

Reference:

Reviews of Modern Physics. Lancaster, P.A.: Published for the American Physical Society by the American Institute of Physics, 1952. Print.

Wehr, M. Russell, and James A. Richards. Physics of the Atom. Reading, MA: Addison-Wesley Pub., 1984. Print.

Young, Hugh D., and Roger A. Freedman. University Physics. Boston, MA: Pearson Custom, 2008. Print.

Adhikari, P.B. A Textbook of Physics. 2070 ed. Vol. II. Kathmandu: Sukunda Publication, 2070. Print.