Question

39. The separation between the lines of a pure rotational molecular spectrum is:
(A)B
(B) 2B
(C) 4B
(D)6B​

Answer 1

Answer:

correct answer is 2B...

Answer 2

Answer:

The separation between the lines of a pure rotational molecular spectrum is: 2B Option (B)

Explanation:

• Rotational spectroscopy which is also called microwave spectroscopy provides information about the absorption or emission of electromagnetic radiation typically in the microwave region of the electromagnetic spectrum.

The rotational spectrum of a diatomic molecule consists of equally spaced lines.

The equation for the frequency separation of the lines in the rotational spectrum of a  diatomic molecule.

The rotational energy EJ of a diatomic molecule is given by,

$\mathrm{EJ}=\frac{h^{2}}{8 \pi^{2} I} J(J+1) -----(1)$

Where, I =moment of inertia of the molecule

h =Planck s constant

J =rotational quantum number which can have value 0,1,2,3……

Consider a rotational transition from a rotational energy level J to another level (J1)

$\text { E } \mathrm{J}^{\prime}=\frac{\mathrm{h}^{2}}{8 \pi^{2} I} J^{\prime}\left(J^{\prime}+1\right)----(2)$(c) Now change its rotation energy (∆E) is given as,

                            ∆E = EJ – EJ’

$\begin{aligned}&\text { } \Delta \mathrm{E}=\frac{\mathrm{h}^{2}}{8 \pi^{2} I} J(J+1)-\frac{\mathrm{h}^{2}}{8 \pi^{2} I} J^{\prime}\left(J^{\prime}+1\right)\\&\text { } \Delta \mathrm{E}=\frac{\mathrm{h}^{2}}{8 \pi^{2} I}\left[J(J+1)-J^{\prime}\left(J^{\prime}+1\right)\right]----(3)\end{aligned}$

But accordingly, to the selection rule, ∆J=+1

                            i.e. ∆J= J-J’= +1

                            Therefore J’= J-1

By substituting equation of J’ in eq (3) we get,

$\begin{aligned}&\Delta \mathrm{E}=\frac{\mathrm{h}^{2}}{8 \pi^{2} I}[J(J+1)-(J-1)(J-1+1)]\\&\text { } \Delta \mathrm{E}=\frac{\mathrm{h}^{2}}{8 \pi^{2} I}\left(\mathrm{~J}^{2}+\mathrm{J}-\mathrm{J}^{2}+\mathrm{J}\right)\\&\text { } \Delta E=\frac{h^{2}}{8 \pi^{2} I} \times 2 J----(4)\end{aligned}$

(d) According to quantum theory, energy changes are quantized by the equation,

           ΔE = hυ = hῡc            ……………  ………………….…. ..(v)

$\begin{aligned}&\mid * \Delta E=\mathrm{h}^{*} \bar{v}_{\mathrm{C}}=\frac{\mathrm{h}^{2}}{8 \pi^{2} I} \times 2 \mathrm{~J} \quad \ldots \ldots \ldots .(\text { from eq (iv) \& (v)) }\\&\text { * } \mathrm{h} \overline{\mathrm{v}}_{\mathrm{c}}=\frac{\mathrm{h}^{2}}{8 \pi^{2} I} \times 2 \mathrm{~J}\\&\text { But, } \frac{\mathrm{h}}{8 \pi^{2} I c}=\mathrm{B} \text { i.e Rotational constant or Bjerrum's constant. }\end{aligned}$

Therefore ῡ = 2BJ

The frequency separation (∆v) between two successive lines in the rotational spectrum is 2B  i.e ∆ v = 2B.