The bond length of HCI in the ground electronic state is 1.27455 A. 1 A = 10^-10

Question

The bond length of HCI in the ground electronic state is 1.27455 A. 1 A = 10^-10 meter. The mass of a 1H nucleus is 1.007825 amu and the mass of a 35CI nucleus is 34.96885 amu. Calculate the moment of inertia of 1H^35Cl. Calculate the rotational partition function for an individual 1H^35Cl molecule, at 300 K. The probability for a 'H^35Cl molecule to be in the rotational level with the angular momentum quantum number J is given by where h = h (2pi), I is the moment of inertia of the molecule, and q_rot is the rotational partition function. Find the probability for a 'H^35Cl molecule to be in the ground rotational state (j = 0) at T = 300 K. Find the probability for a 1H^35Cl molecule to be in the level with J = 2at 300 K. 1H^37Cl has the same bond length as 1H^35C1 to a very good approximation, but the mass of a 37Cl nucleus is 36.96590 amu. Compare the probability to find iH^37Cl in the ground rotational state (J = 0) with the probability you found for 1H^35CI in part c.

Explanation / Answer

a) we know that moment of inertia for this type of rigid rotor I= reduced mass (meu)*bond length^2

meu=m1*m2/m1+m2 = 1.007825*34.96885/1.007825+34.96885 =35.24/35.97 =.9797amu

now bond length is given R= 1.27455 Angstrom

then I=R^2= .9797*1.27455 ^2 =1.591 units

b) The rotational partition function qrot= KT/B

where = symmetry numer = 2

B= h/8pie^2Ic = .052*10^-34/c

Kt=KbT/hc = 62.53*10^11/c

using all the values we get qrot= 601.25*10^8

c) probability at J=0 and T=300K

Pj= (2.0 +1)exp(0/qrot) =1

d)again Pj at j=2 and T=300k

Pj= (2.2 +1)exp(-2(2+1)hcut^2/2.1.591.1.38*10^-23.300)/601.25*10^8

we use hcut as reduced plank constsnt hcut=h/2pie

e) at J=0, probability is independent of mass in this case

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