Student Solutions Manual for Ball's Physical Chemistry, 2nd
Student Solutions Manual for Ball's Physical Chemistry, 2nd
2nd Edition
ISBN: 9798214169019
Author: David W. Ball
Publisher: Cengage Learning US
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Chapter 16, Problem 16.15E
Interpretation Introduction

Interpretation:

The ΔE values experienced by the ground state (4F3/2) energy levels of V atom are to be calculated.

Concept Introduction:

The change in energy (ΔE) of the state depends on the strength of the magnetic field (B) and the z-component quantum number (ML) in the presence of magnetic field. The expression for the change in energy (ΔE) is written below.

ΔE=μBMLB

The value of (ML) can be positive, negative or zero. Hence, the change in energy (ΔE) can be positive, negative or zero. The change in energy (ΔE) of the state is equal to the amount of splitting.

Expert Solution & Answer
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Answer to Problem 16.15E

The ΔE values experienced by the ground state (4F3/2) energy levels of V atom are 3.10×1024J, 1.03×1024J, +1.03×1024J and +3.10×1024J.

Explanation of Solution

The ΔE values experienced by the ground state (4F3/2) energy levels of V atom are calculated by the formula given below.

ΔE=gμBMJB…(1)

Where,

μB is Bohr magneto (9.274×1024J/T).

(MJ) is the z-component quantum number.

g is Lande g factor.

B is earth’s magnetic field.

The value of g is calculated by the formula given below.

g=1+j(j+1)+s(s+1)l(l+1)2j(j+1)…(2)

Where,

j, s and l are quantum numbers.

The values of j, s and l for (4F3/2) state are 3/2, 3/2 and 3.

Substitute the values of j, s and l in equation (2).

g=1+3/2(3/2+1)+3/2(3/2+1)3(3+1)2×3/2(3/2+1)=1+15/4+15/41230/4=1+30/41230/4=118/430/4

Simplify the above equation.

g=135=25

One gauss equals to 1.0×104T. Therefore 5.57×103G will be equal to 0.557T.

The value of μB and B are (9.274×1024J/T) and 0.557T.

Substitute the values of μB, B and g in equation (1).

When MJ=(32),

ΔE=gμBMJB=25×9.274×1024J/T×(32)×0.557T=(35)×5.165×1024J=3.10×1024J

When MJ=(12),

ΔE=gμBMJB=25×9.274×1024J/T×(12)×0.557T=(15)×5.165×1024J=1.03×1024J

When MJ=0, ΔE will be zero.

When MJ=32,

ΔE=gμBMJB=25×9.274×1024J/T×32×0.557T=35×5.165×1024J=3.10×1024J

When MJ=12,

ΔE=gμBMJB=25×9.274×1024J/T×12×0.557T=15×5.165×1024J=1.03×1024J

Conclusion

The ΔE values experienced by the ground state (4F3/2) energy levels of V atom are 3.10×1024J, 1.03×1024J, +1.03×1024J and +3.10×1024J.

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