Calculate the energies of the first four rotational levels of 1 H127I free to rotate in...
Calculate the energies of the first four rotational levels of 1 H127I free to rotate in three dimensions, using for its moment of inertia I=µR2 , with µ=mHmI /(mH+mI ) and R=160 pm.
Homework Answers
Reduced mass μ = mHmI/(mH+mI)
= (1 x 127)/(1 + 127) x 1.661 x 10-27 = 1.648 x 10-27 kg
Momemt of inertia I = μR2
= 1.648 x 10-27 x (160 x 10-12)2 = 4.219 x 10-47 kg m2
Rotational constant B = h2/8π2I
= (6.626 x 10-34)2/(8 x 3.14162 x 4.219 x 10-47) = 1.32 x 10-22 J
Rotational energy level: EJ = BJ(J + 1)
First four rotational levels are:
E0 = 1.32 x 10-22 x 0 x 1 = 0 J
E1 = 1.32 x 10-22 x 1 x 2 = 2.64 x 10-22 J
E2 = 1.32 x 10-22 x 2 x 3 = 7.92 x 10-22 J
E3 = 1.32 x 10-22 x 3 x 4 = 1.58 x 10-21 J
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