The vibrational energy levels, v11 = 0 and v1 = 1 of a diatomic molecule are separated by 2143 cm-1. Its Anharmonicity (ωe Xe ) is 14 cm-1. The values of ωe (in cm-1) and first overtone (cm-1) of this molecules are respectively.a)2171, 4300b)2157, 4286c)2157, 4314d)2171, 4258Correct answer is option 'A'. Can you explain this answer?

Question

The vibrational energy levels, v11 = 0 and v1 = 1 of a diatomic molecule are separated by 2143 cm-1. Its Anharmonicity (&omega;e Xe ) is 14 cm-1. The values of &omega;e (in cm-1) and first overtone (cm-1) of this molecules are respectively.a)2171, 4300b)2157, 4286c)2157, 4314d)2171, 4258Correct answer is option 'A'. Can you explain this answer?

Answer

The deviation from the harmonic oscillator model) is 100 cm-1.

To find the anharmonicity constant, we can use the formula:

ω = ω_e - ω_ex_e(v + 1/2) - ω_ey_e(v + 1/2)^2 - ω_ez_e(v + 1/2)^3

where:
ω is the vibrational frequency of the molecule
ω_e is the equilibrium vibrational frequency
ω_ex_e is the vibrational constant (force constant)
ω_ey_e is the anharmonicity constant

We know from the problem that:

ω_e = (v11 + v1) / 2ω = 2143 cm-1
v11 = 0
v1 = 1

Substituting these values into the formula, we get:

2143 = ω_e - ω_ex_e(1/2) - ω_ey_e(3/4) - ω_ez_e(5/8)

Since we are only interested in the anharmonicity constant, we can ignore ω_e, ω_ex_e, and ω_ez_e:

- ω_ey_e(3/4) = -100 cm-1

Solving for ω_ey_e, we get:

ω_ey_e = 133.33 cm-1

Therefore, the anharmonicity constant for this diatomic molecule is 133.33 cm-1.

Answer

The value of Xe is not 14. Correct answer should be 4258.

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