Energy eigen value of Rigid rotator

Energy Eigen Value of Rigid rotator:

From equation (16) and (21)

$$\beta=\frac{2IE}{\hbar^2}=l(l+1)$$

$$\therefore\; E= \frac{\hbar l(l+1)}{2I}\dotsm(27)$$

It gives the energy eigen values of rigid rotator.

Equation (27) can be written in the form,

$$E_l= Bh l(l+1)\dotsm(28)$$

Where, \(B=\frac{h}{8\pi^2 I} \) ( Rotational constant )

$$B= \frac{h}{8\pi^2\mu r^2}$$

The value of l is given by l=0,1,2,3,...

$$E_0= 0$$

$$l=1, \; E_1= 2Bh$$

$$l=2,\; E_2= 68Bh$$

$$l=3,\; E_3 = 12 Bh$$

$$l=4,\; E_4= 20Bh$$

$$l=5,\; E_5= 30Bh$$ and so on.

1. Energy eigen values of rigid rotator is discrete and energy is quantised.

2. Each energy level is (2l+1) fold degenerate. [ i.e. there are (2l+1) wave function for each value of l and \(E_l\))

3. The different wave function for each value of l and indicated by different values of m. For example For, l=3. there are 7 wave function having same energy \(E_3= 12 Bh\). They are symbolised as,

$$\psi_l^m (\theta, \phi); [\psi_3^{-3}, \psi_3^{-2}, \psi_3^{-1}, \psi_3^{0}, \psi_3^1,\psi_3^2, \psi_3^3]$$

4. The separation between successive energy level increases linearly with quantum number l.

$$i.e \; E_{l+1}-E_l= Bh[ (l+1)((l+2)-l(l+1)]$$

$$=Bh(l^2+ 3l+ 2-l^2-1)$$

$$=2Bh(l+1)$$

$$\Rightarrow (E_{l+1}-E_l)\propto l+1$$

$$\Rightarrow (E_l-E_{l-1})\propto l$$

5. The energy gap between successive level is not equal. ( Gap increases as we move towards higher energy level).

6. The probability density corresponding to \(\psi_1^m(\theta, \phi)\) is always independent of \(\phi\). [ \(\phi\)= azimuthal angle]

i.e Probability distribution is always symmetrical about z-axis as,

$$\phi(\theta,\phi)= [\psi_l^m (\theta, \phi)]^*[\psi_l^m (\theta, \phi)]$$

$$=\biggl[\frac{2l+1}{4\pi}\frac{(l-m)!}{(l+m)!}\biggr] |P_l^m (cos\theta)^2| e^{-im\phi}\cdot e^{im\phi}$$

$$=\biggl[\frac{(2l+1)}{4\pi}\frac{(l-m)!}{(l+m)!}\biggr]|P_l^m (cos\theta)|^2= independent \; of \phi$$

$$= Symmetric \; about \; z-axis.$$

7. The energy level diagram of rigid rotator is given by;

Question : If Bound length of Hcl molecule is \(1.3 \; A^0\). Then determine the 1st four rotational energy levels of the molecules in terms of electron volt (ev).

\(\Rightarrow\) Soltuion:

Bond length of Hcl molecule (r) = \( 1.3^\circ= 1.3\times 10^{-10}m\)

mass of H-atom \( ( m_H)= 1\;amu\)

mass of chlorine atom \((m_{cl})= 35.5\; amu\)

$$\mu_{Hcl}= \frac{m_H\cdot m_{cl}}{m_H+ m_{cl}}$$

$$=\frac{1\times 35.5}{1+ 35.5}$$

$$=\frac{35.5}{36.5}$$

$$=0.97\; amu$$

$$=0.97\times 1.67\times 10^{-27}kg$$

Now,

$$B=\frac{h}{8\pi^2\mu_{Hcl}r^2}$$

$$=\frac{6.62\times 10^{-34}}{8\times (3.14)^2\times 0.97\times 1.67\times 10^{-27}\times (1.3\times 10^{-10})^2}$$

$$=3.06\times 10^{11}sec^{-1}$$

Now, Rotational energy of Hcl molecule is,

$$E_l= Bh(l+1)\; For l= 0,1,2,3,..$$

$$For\; l=0, E_0=0$$

$$For \; l=1, E_1= 2Bh= 2\times 3.06\times 10^{11}\times 6.62\times 10^{-34}$$

$$=4.05\times 10^{-22}J$$

$$=\frac{4.05\times 10^{-22}}{1.602\times 10^{-19}}ev$$

$$=2.62\times 10^{-34}ev$$

$$For\; l=2, E_2= 6Bh=6\times 3.06\times 10^{11}\times 6.62\times 10^{-34}$$

$$=7.58\times 10^{-3}ev$$

$$For,\;l=3, E_3= 12Bh= 12\times 3.06\times 10^{11}\times 6.62\times 10^{-34}$$

$$=0.015ev$$

Reference:

1. Mathews, P.M and K Venkatesan.A Text Book of Quantum Mechanics.New Delhi: Tata McGraw Hill Publishing Co. Ltd, 1997.
2. Merzbacher, E.Quantum Mechanics .New York: John Wiley, 1969.
3. Prakash, S and S Salauja.Quantum Mechanics.Kedar Nath Ram Nath Publishing Co, 2002.
4. Singh, S.P, M.K and K Singh.Quantum Mechanics.Chand & Company Ltd., 2002.