Answer to Question #122605 in Quantum Mechanics for komal

Question

For a carbon monoxide, CO molecule the energy of a transition from the â„“ =2 to the â„“ = 3 state is 1.43 x 10-3 eV.

i) Compute the rotational inertia of the CO molecule.

ii) What is the average separation between the centers of the C and O atoms?

Answer 1

When a wave travels in an electromagnetic field covers a certain distance while completing the wave cycles. The wavenumber tells the number of cycles which are achieved under the particular distance. It is different from the frequency because the frequency is associated with time while the wavenumber is related to distance.

Solution and Explanation:

The energy in transition is:

E = 1.43 x 10^-3 eV = 1.43 x 10^-3 x 1.602 x 10^-19 = 2.3 x 10^-22 J

(a) The expression to calculate the rotational inertia is given as:

$ln = (h/8\pi^2 Ic ) (n+1) =\nu$

$I = n(h/8π^2νc) (n+1)$

Here, n is the state, h is Planck's constant, ν is the wave number and c is the speed of light.

The standard value of Planck's constant is 6.626 x 10^-34 J*s

The standard value of speed of light is 3 x 10⁸ m/s

Transition state is 2.

Calculate the wave number:

v = E/hc

v = 2.3 x 10^-22 / 6.626 x10^-34x 3 X 10⁸

v = 0.1157 x 10⁴ m^-1

Substituting the value in the above expression, we get:

I = 2* (6.626 x 10^-34 / 8 x[3.14]² x 0.1157 x 10⁴ x 3 x 10⁸) (2+1)

I = 1.42 x 10-46 kg*m2

(b) The expression to calculate the separation between the C and O atoms is:

$r = \sqrt \frac {I}{m_\alpha}$

Here, m_α is the average mass.

m_α = m_cm_o/m_c+m_o

m_α = 1.994 x 2.656 x 10^-52 / (1.994 + 2.656) x 10^-26

m_α = 1.139 x 10^-26 kg

$r = \sqrt \frac {1.42 x 10^-46}{1.139 x 10^-26}$

$r = \sqrt{1.24 \Chi10^-26}$

r = 1.11 x 10^-10 m

Hence, the separation between the C and O atoms is 1.11 x 10^-10 m