Question

The inter atomic distance of 14N16O molecule is 115.1 pm. Calculate (10)
(i) its reduced mass,
(ii) its moment of inertia,
(iii) the wave number of the line corresponding to lowest absorption in m1 Unit, and
(iv) the energy in m1 unit for the transition J = 2 to J = 3.

Answer 1

Answer on Question #54475, Chemistry / General chemistry

Question:

The interatomic distance of $^{14}\mathrm{N}^{16}\mathrm{O}$ molecule is 115.1 pm. Calculate

(i) its reduced mass,

(ii) its moment of inertia,

(iii) the wave number of the line corresponding to lowest absorption in $\mathrm{m}^{-1}$ Unit, and

(iv) the energy in $\mathrm{m}^{-1}$ unit for the transition $J = 2$ to $J = 3$.

Solution:

(i) The mass of a nitrogen atom is 14.003 amu;

the mass of an oxygen atom is 15.995 amu;

and the conversion factor is $1.6605 \times 10^{-27} \, \mathrm{kg/amu}$.

The reduced mass is

$$\mu = \frac{\mu_N \mu_O}{\mu_N + \mu_O}$$ $$\mu = \frac{14.003 \times 15.995}{14.003 + 15.995} = 7.4664 \, \mathrm{amu} = 7.4664 \times 1.6605 \times 10^{-27} \, \mathrm{kg} = 1.24 \times 10^{-26} \, \mathrm{kg}$$

(ii) The moment of inertia is

$$I = \mu R^2 = 1.24 \times 10^{-26} \times (115.1 \times 10^{-12})^2 = 1.64 \times 10^{-46} \, \mathrm{kg \cdot m^2}$$

(iii) The rotational constant is

$$\bar{B} = \frac{h}{8\pi^2 c I}$$ $$\bar{B} = \frac{6.626 \times 10^{-34} \, \mathrm{J \, s}}{(8\pi^2)(2.998 \times 10^8 \, \mathrm{m/s})(1.64 \times 10^{-46} \, \mathrm{kg \, m^2})} = 170.6 \, \mathrm{m^{-1}}$$

Since the energy at which each "line" is measured is given by $E_J - E_J$, the shortest line or lowest energy transition occurs for $J = 0 \rightarrow J = 1$;

Wavenumber is

$$F = B J'(J' + 1) - B J(J + 1) = B(1(1 + 1) - 0(0 + 1)) = 2B = 2 \times 170.6 = 341.2 \, \mathrm{m^{-1}}$$

(iv) $E = B J'(J' + 1) - B J(J + 1) = B(3(3 + 1) - 2(2 + 1)) = 6B = 6 \times 170.6 \, \mathrm{m^{-1}}$

$$E = 1023.6 \, \mathrm{m^{-1}}$$

www.AssignmentExpert.com