Question

Rigid rotor wavefunctions are given by \( Y_{l,m}(\theta, \phi) \). The wavefunctions \( Y_{1,0}(\theta, \phi) \) and \( Y_{2,0}(\theta, \phi) \) are given below: \[ Y_{1,0}(\theta, \phi) = \sqrt{\frac{3}{4\pi}} \cos \theta \qquad Y_{2,0}(\theta, \phi) = \sqrt{\frac{5}{16\pi}} (3\cos^2 \theta - 1) \] For a non-polar diatomic molecule, the value of transition dipole moment integral for the transition between \( Y_{1,0}(\theta, \phi) \) and \( Y_{2,0}(\theta, \phi) \) is equal to:

• A. \( \frac{1}{\sqrt{2\pi}} \)
• B. \( 0 \)
• C. \( 2 \)
• D. \( \frac{1}{\sqrt{4\pi}} \)

Hint:

Transition dipole moment integrals vanish between certain spherical harmonics due to their orthogonality and parity properties. For non-polar molecules, transitions between \( Y_{1,0} \) and \( Y_{2,0} \) are forbidden.

Answer 1

The Correct Option is B

The transition dipole moment integral is given by: \[ \mu = \int Y_{2,0}^*(\theta,\phi) \, \mu \, Y_{1,0}(\theta,\phi) \, \sin \theta \, d\theta \, d\phi \] Since the molecule is non-polar and the dipole lies along the \( z \)-axis, only the \( z \)-component contributes: \[ \mu_z \int_0^\pi \int_0^{2\pi} Y_{2,0}^*(\theta,\phi) \cos \theta \, Y_{1,0}(\theta,\phi) \, \sin \theta \, d\phi \, d\theta \] Using the given wavefunctions: \[ Y_{1,0} = \sqrt{\frac{3}{4\pi}} \cos \theta, \quad Y_{2,0} = \sqrt{\frac{5}{16\pi}} (3\cos^2 \theta - 1) \] We substitute and evaluate the integral: \[ \int_0^{2\pi} d\phi \int_0^\pi \left[\sqrt{\frac{5}{16\pi}} (3\cos^2 \theta - 1)\right] \cos \theta \left[\sqrt{\frac{3}{4\pi}} \cos \theta\right] \sin \theta \, d\theta \] This simplifies to: \[ \sqrt{\frac{15}{64\pi^2}} \cdot 2\pi \int_0^\pi (3\cos^2 \theta - 1) \cos^2 \theta \sin \theta \, d\theta \] Now simplify the integrand: \[ (3\cos^2 \theta - 1) \cos^2 \theta \sin \theta = (3\cos^4 \theta - \cos^2 \theta) \sin \theta \] This function is odd in terms of \(\cos \theta\) over the interval \( [0, \pi] \), so the total integral is zero. \[ \boxed{0} \]