Sommerfield atomic model: Calculation of total energy

Total energy of electron in non-relativistic Sommerfield atomic model:

Total energy is the sum of kinetic and potential energy,

$$i.e\;\; E_n= K.E + P.E $$

$$= \frac 12 \biggl[ \biggl(\frac{dr}{dt}\biggr)^2+\biggl(\frac{d\phi}{dt}\biggr)^2\biggr] + \frac{1}{4\pi} \frac{(+ze)(-e)}{r}\dotsm(1)$$ Where ,

\(\frac {dr}{dt}\)= radial velocity

\(\frac{d\phi}{dt}\)= angular velocity

$$ p_r = m \frac {dr}{dt}\Rightarrow \frac {dr}{dt}= \frac {p_r}{m}\dotsm(2)$$ $$p_\phi= mr^2\frac {d\phi}{dt}\Rightarrow\frac{d\phi}{dt}=\frac{ p_\phi}{mr^2}$$

Using (2) in (1),

$$K.E= \frac 12 m \biggl[ \frac{p_r^2}{m}+ \biggl(\frac{rp_\phi}{mr^2}\biggr)^2\biggr]$$

$$= \frac {1}{2m}[pr^2+\frac{1}{r^2}p_\phi^2]$$

$$=\frac{p_\phi^2}{2mr^2}\biggl[\frac{r^2p_r^2}{p_\phi^2}+1\biggr]$$

$$=\frac{p_\phi^2}{2mr^2}\biggl[\frac{r^2}{p_\phi^2}(\frac{dr}{dt})^2+1\biggr]$$

We have,

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

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

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

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

$$ or, p_r d_r= \biggl(\frac{\epsilon sin\phi}{1+cos\phi}\biggr)^2 p_\phi d\phi\dotsm(4)$$

$$ or, \frac 1r= \frac{1+\epsilon cos\phi}{0+1(1-\epsilon^2)}\dotsm(5)$$

From equation (3)

$$ K.E = \frac{p_\phi^2}{2m}\biggl[\frac{r^2(\frac{dr}{dt})^2m^2}{p_\phi^2}+1\biggr]$$

$$=\frac{p_\phi^2}{2mr^2}\biggl[\frac{r^2(\frac{dr}{d\phi}\frac{d\phi}{dt})^2m^2}{p_\phi^2}+1\biggr]$$

$$=\frac{p_\phi^2}{2mr^2}\biggl[\frac{r^2(\frac{dr}{dt}\frac{p_\phi}{mr^2})^2m^2}{p_\phi^2}+1\biggr]$$

$$=\frac{p_\phi^2}{2mr^2}\biggl[\biggl(\frac{1}{r} \frac{d\phi}{dt}\biggl)^2+1\biggl]$$

$$=\frac{p_\phi^2}{2m}\frac{-(\epsilon sin\phi)}{1+cos\phi}\frac{(1+\epsilon cos\phi)^2}{a^2(1-\epsilon^2)}\biggl[\biggl(\frac{\epsilon sin\phi}{1+\epsilon cos\phi}\biggl)^2\biggl]$$

$$=\frac{p_\phi^2}{2m}\frac{(1+\epsilon cos\phi)^2}{a^2(1-\epsilon^2)^2}\biggl[\frac{\epsilon^2 sin^2\phi+1+2\epsilon cos\phi+\epsilon^2 cos^2\phi}{(1+\epsilon cos\phi)^2}\biggr]$$

$$=\frac{p_\phi^2}{2m}\frac{(1+\epsilon cos\phi)^2}{a^2(1-\epsilon^2)^2}\biggl[\frac{\epsilon^2+1+2\epsilon cos\phi}{(1+\epsilon cos\phi)^2}\biggr]$$

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

Again,

$$P.E= -\frac{1}{4\pi\epsilon_circ} ze^2\cdot\frac 14$$

Using (5),

$$P.E= -\frac{1}{4\pi\epsilon_\circ} ze^2\cdot\frac{1+\epsilon cos\phi}{a(1-\epsilon^2)}\dotsm(7)$$

Total energy ,

$$T.E= \frac{p_\phi^2}{2m}\frac{1}{a^2(1-\epsilon^2)^2}\cdot (1+\epsilon^2+2\epsilon cos\phi) - \frac{1}{4\pi\epsilon_circ}\cdot \frac{ze^2(1+\epsilon cos\phi)}{a(1-\epsilon^2)}$$

$$or, T.E= \biggl[ \frac{p_\phi^2(1+\epsilon^2)}{2ma^2(1-\epsilon^2)^2}-\frac{ze^2}{4\pi\epsilon_\circ(1-\epsilon^2)}\biggr]+\biggl[\frac{p_\phi^2\epsilon}{ma^2(1-\epsilon^2)^2}-\frac {ze^2\epsilon}{4\pi\epsilon_\circ}\frac{\epsilon}{a(1-\epsilon^2)}\biggr]cos \phi\dotsm(8)$$

Due to presence of \( cos\phi\) in the expression o ftotal energy, total energy becomes variable.

According to conservation of energy total energy should be constant. ie the coefficient of \(\cos\phi\) must be zero.

$$i.e\;\; \frac{p_\phi^2 \epsilon}{ma^2(1-\epsilon^2)^2}-\frac{ze^2\epsilon}{4\pi\epsilon_\circ a(1-\epsilon^2)}=0$$

$$or, \frac{\epsilon}{a(1-\epsilon^2)}\biggl[\frac{p_\phi^2}{a(1-\epsilon^2)}-\frac{ze^2}{4\pi\epsilon_\circ}\biggr]=0$$

$$\Rightarrow \frac{p_\phi^2}{ma(1-\epsilon^2)}-\frac{ze^2}{4\pi\epsilon_\circ}=0$$

$$\Rightarrow \frac{p_\phi^2}{ma(1-\epsilon^2)}-\frac{ze^2}{4\pi\epsilon_\circ}\dotsm(9)$$

Since\(\frac{\epsilon}{a(1-\epsilon^2)}\neq0\)

$$\Rightarrow a=\frac{p_\phi^2 4\pi\epsilon_\circ}{m(1-\epsilon^2)ze^2}$$

$$=\frac{(\frac{n_\phi h}{2\pi})^2 4\pi\epsilon_\circ}{m(\frac{n\phi}{n})^2}$$

$$=\frac{\frac{h^2}{4\pi^2}4\pi\epsilon_\circ n^2}{m}$$

$$\Rightarrow\frac{\epsilon_\circ n^2 h^2}{\pi m ze^2}= radius\;of\;1^st\;Bhor's\;orbit\;of\;H-atom\dotsm(10)$$

Again using (9) in equation (8)

$$E=\biggl[\frac{p_\phi^2(1+\epsilon^2)}{2ma^2(1-\epsilon^2)^2}-\frac{Ze^2}{4\pi\epsilon_\circ a^2(1-\epsilon^2)}\biggr]$$

$$=\;\;\frac{p_\phi^2}{ma(1-\epsilon^2)}\frac{1+\epsilon^2}{2a(1-\epsilon^2)}-\frac{ze^2}{4\pi\epsilon_\circ a^2(1-\epsilon^2)}$$

$$=\:\:\frac{ze^2}{4\pi\epsilon_\circ}\frac{1+\epsilon^2}{2a(1-\epsilon^2)}-\frac{2e^2}{4\pi\epsilon_\circ a^2(1-\epsilon^2)}$$

$$=\;\;\frac{ze^2}{4\pi\epsilon_\circ a(1-\epsilon^2)} \biggl(\frac{1+\epsilon^2}{2}-1\biggr)$$

$$=\:\frac{-ze^2}{4\pi\epsilon_\circ a(1-\epsilon^2).2} (1-\epsilon^2)$$

$$=\;\;\frac{-ze^2}{4\pi\epsilon_\circ 2a}$$

$$=\;\; \frac{-ze^2}{8\epsilon_\circ \pi}\cdot\frac{\epsilon_\circ n^2 h^2}{\pi mze^2}\;\;\;\; [ using (10)]$$

\(E_n=\frac{-me^4Z^2}{8\epsilon_\circ^2n^2h^2}\)= Energy given by Bohr's theory. \(\dotsm\)(11)

Non- relativistic Sommerfield atomic model gives the same energy as in Bohr's theory so it can not explain the splitting of \(H_\alpha,\;H_\beta,\;H_\gamma\) lines.

Failure of Sommerfield atomic model:

1. It couldn't explain the proper number of fine structure lines in spectrum of H-atom.

2. It could not explain the complex spectra of multi-electron atom.

3. It couldn't explain the complex spectra of multi-electron atom.

4. It couldn't explain the distribution and motion of electrons in atom.

6. It could not explain 'Zeeman effect' and 'Strak effect'

So, to explain above inadequacies of Sommerfield atomic model a new atomic model is proposed known as 'vector model atom'.

Reference:

1. Agrawal, Fundamentals of Modern Physics, Pragati Prakashan Meerut -26000 (U.P).

2. G Aruldhas and P.Raja Gopal, Modern Physics, Prentice-Hall of India Private Limited, New Delhi -11001.

3. Gupta, Satish K. and Pradhan, J.M (2005), A Textbook of Physics, Part - I & II, Surya Publications, Jalandhar City, India.

4.. D.C. Tayal, Nuclear physics, Himalaya publishing house, Ramdoot, Dr. Bhalerao Marg, Bombay, 400 004.