The moments of Inertia for NH_3 are I_b = I_c = 2.816 times 19^-47 kg-m^2 and I_

Question

The moments of Inertia for NH_3 are I_b = I_c = 2.816 times 19^-47 kg-m^2 and I_a = 4.43 times 10^-47 kg m^2. Compute the rotational energy states for J = 0, 1 and 2 with all allowed values of K. Suppose we promote the NH_3 molecule from the J = 0, K = 0 to the J = 1, K = 1 quantum state. Calculate the wavelength of light required to do this. Compute the moments of Inertia for ND_3 where D is the deuterium atom. Repeat b.) for the ND_3 molecule. Can we use the wavelength to determine which molecule is present? What would happen if the molecule were now NH_2D? Can we solve this system as we did NH_3 and ND_3? If so, solve the system for the wavelength. If not, explain why.

Explanation / Answer

Solution 5:-

a) By Schrodinger equation, we know that

E(J)=B J(J+1)

So rotational energy states are follows:-

1) E(0) = B 0 (0+1)

E(0) = 0

2) E(1) = B 1 (1+1) = B 1 (2) and

3) E (2) = B 2 (2+1) = B 1 (3)

Here in all cased B = h/ 8 * pi()^2 cl

Solution b) Suppose we promote the NH3 molecules from J=0 k = 0 to J=1 k = 1 quantum state. Then the wavelength of light to do this varies with respect to the square root of rotational energy states

Solution c):- The moment of inertia for ND3 where D is deuterium ion about an axis is shown below:-

= r^2. dm

Here

dm = an element of mass of the ND3

and r is basically the perpendicular distance from the axis to any arbitrary mass element of the body.

Solution d) It will be exactly same as part b and yes we can use wavelength to determine which molecule is present.

Solution e) If the molecule were now NH2D then we cannot solve this system because of two different elements H and D present in the system which cannot be solved separately.

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