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The rovibrational transition of^1H^35Cl with v = 0 rightarrow 1, J = 11 rightarrow 10 occurs at 2757.89 cm^-1, and the transition with v = 0 rightarrow 1, J = 10 rightarrow 9 occurs at 2779.07 cm^-1. From
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The rovibrational transition of^1H^35Cl with v = 0 rightarrow 1, J = 11 rightarrow 10 occurs at 2757.89 cm^-1, and the transition with v = 0 rightarrow 1, J = 10 rightarrow 9 occurs at 2779.07 cm^-1. From this information,

i) calculate the spring constant of the vibrational potential (assuming the harmonic approximation and rigid rotor approximation) and
ii) the equilibrium length of the HCl bond.

User CorreyS
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Answer:

Step-by-step explanation:

find the attached solution below

The rovibrational transition of^1H^35Cl with v = 0 rightarrow 1, J = 11 rightarrow-example-1
The rovibrational transition of^1H^35Cl with v = 0 rightarrow 1, J = 11 rightarrow-example-2
User Marduk
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