Question 520: Atoms & Nuclei

Question

Using Bohr's quantisation condition, what is the rotational energy in the second orbit for a diatomic molecule? ($$I=$$ moment of inertia of diatomic molecule and, $$h=$$ Planck's constant)

Accepted Answer

The angular momentum of diatomic molecules,
$$L=I \omega=\frac{n h}{2 \pi}$$
For second orbit, $$L=\frac{2 h}{2 \pi}=\frac{h}{\pi}$$
Rotational energy
$$=\frac{1}{2} I \omega^2=\frac{(I \omega)^2}{2 I}=\frac{L^2}{2 I}=\left(\frac{h}{\pi}ight)^2 	imes \frac{1}{2 I}=\frac{h^2}{2 h \pi^2}$$

Answer

$$\frac{h}{2 I \pi^2}$$

Answer

$$\frac{h^2}{2 I \pi^2}$$

Answer

$$\frac{h^2}{2 I^2 \pi^2}$$

Answer

$$\frac{h}{2 I^2 \pi}$$

Atoms & Nuclei

Using Bohr's quantisation condition, what is the rotational energy in the second orbit for a diatomic molecule? ($I= moment of inertia of diatomic molecule and, h=$ Planck's constant)

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