Quantum Chemistry

Chapter10 - Home
Exercises: 1 - 2 - 3 - 4 - 5 - 6 - 7 - 8 - 9 - 10 - 11 - 12 - 13 - 14 - 15 - 16 - 17 - 18 - 19 - 20 - 21 - 22 - 23 - 24 - 25 - 26 - 27 - 28 - 29 - 30 - 31 - 32 - 33 - 34 - 35 - 36 - 37 - 38 - 39 - 40 - 41 - 42 - 43 - 44 - 45

Exercise. This is the solution to exercise 10.1 in the book.

Solution. From

EJ = 2 2IJ(J + 1)

we get

ΔE = 2 I (J + 1)

or, replacing ΔE we get

hν = h2 4π2I(J + 1)

or for J = 0

I = h 4π2ν

Replacing with numerical values we get:

I = 2.643 1047kgm2

For this molecules

μ = 1 35 1 + 35 1.67265 1027kg = 1.6262 1027kg

Since

I = μR02

then

R0 = 1.275Å