Question

Molecular Rotations a. The wavefunction of rotations of diatomic molecules according to the rigid rotor approximation are spherical harmonics. Where have you seen spherical harmonics before? What are the quantum numbers that specify the wavefunctions for the rotational quantum states of a diatomic molecule? b. What are the gross and specific selection rules for pure rotational spectroscopy of a diatomic molecule? What region of the spectrum is used spectroscopy? What are the rotational energy levels for diatomic molecules and spherical rotors? How are these related to the rotational term, F(J)? Show that the c. B-_"_-, has unites of wavenumbers. d. If the wave number of the J-3 2 trannsition of H35Cl considered as rigid rotor is 63.56 cm1, what is i. The moment of inertia of the molecule ii. The bond length?

Answer 1

a. Spherical Harmonics can be seen as two masses m1 amd m2 rotating about their center of mass.

The quantum number that specify the wavefunctions for the rotational quantum state of diatomic molecule is angular quantum number with lowest possible value of 0.

b.The gross selection rule for pure rotational spectroscopy is that the molecule must have permanent dipole moment ,i.e, the molecule must be polar.

The specific selection rule for pure rotational spectroscopy is,

?K = 0, ?J = ±1 where J is rotational quantum number and K has values from -J to +J.

The region for spectrum used is microwave region.

c.

$B = \frac{h}{4\\Pi cl}$

unit of h(planks constant) = Joules-second= kg-m²/s

unit of c(speed of light) = Metre/second

unit of I(Moment of inertia) = Kg-Metre²

unit of B = (kg-m²/s)/(Kg-metre² x metre/second)

             = (metre)^-1

So in cgs units it will be cm^-1 which are the units of wavenumber.

d. wavenumber = 2B(J+1) where l is the lower value of J

$63.56 = 2\times B \times 3$

$B = \frac{63.56}{6}$

B = 10.6 cm^-1

$B = \frac{h}{4\\Pi cl}$

$I = \frac{h}{B\times 4\times \Pi \times c}$

$I = \frac{6.626\times 10^{-34}}{1060\times 4\times \Pi \times 3 \times 10^{8}}$

I = 1.659 X 10^-46 kg-m²

I = mr²

$m = \frac{m_{1}\times m_{2}}{m_{1}+m_{2}}$

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

m= 0.9726

$r = \sqrt{\frac{I}{m}}$

$r = \sqrt{\frac{1.6589\times 10^{-46}}{0.9726}}$

r = 1.305 x 10^-23 m