1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Aniruddha Deb Dec 5 '19 at 8:41
  • $\begingroup$ thank for simply restating my question and not answering it at all. $\endgroup$ – Harley McFarlen Dec 5 '19 at 14:29
  • $\begingroup$ How is the energy related to B? $\endgroup$ – Buck Thorn Dec 5 '19 at 17:54
  • 3
    $\begingroup$ 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. $\endgroup$ – Ivan Neretin Dec 5 '19 at 21:13
1
$\begingroup$

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 $$E = \frac{1}{2} I \omega^2 = \frac{L^2}{2I}.$$

So if you formulate your energy in terms of the angular momentum of your rotating particle, you arrive at the inverse relation.

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 $B$.

| improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.