# Rotational Energy Levels

Why does the rotational constant B decrease and transition spacings decrease as the mass of a particle increase?

I understand from a purely equation perspective that since

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

that as $$I$$ increases the denominator increases and so $$B$$ decreases. But what is the physical reasoning behind this? Why or in what way is the rotational constant dependent on mass?

• I is the moment of inertia of the molecule, which is given by $$I = \mu R^2$$ R is the distance between the two atoms and $\mu$ is the reduced mass of a bimolecular system, given by $$\mu = \frac {m_1 m_2} {m_1 + m_2}$$ If the mass of either particle increases, then the reduced mass increases, causing $I$ to increase, which then causes $B$ to decrease. – Aniruddha Deb Dec 5 '19 at 8:41
• thank for simply restating my question and not answering it at all. – Harley McFarlen Dec 5 '19 at 14:29
• How is the energy related to B? – Buck Thorn Dec 5 '19 at 17:54
• See, it is pretty much the same with any quantum system (think of PIB, think of HO). A heavier particle means more classic-like behavior, which means "less discrete" energy spectrum, which means smaller transition spacings. – Ivan Neretin Dec 5 '19 at 21:13

First, take a look at classical physics. The angular momentum of a particle rotating in a plane is defined as $$L = I \omega$$ and its kinetic energy is $$E = \frac{1}{2} I \omega^2 = \frac{L^2}{2I}.$$
In analogy to the classical picture, the eigenvalues of the rotational Schrödinger equation, $$E = hcBJ(J+1),$$ likewise depend quadratically on the angular momentum quantum number, and thus have a similar inverse dependence on the moment of inertia through $$B$$.