Potential energy of two diatomic molecules P and Q of the same reduced mass is shown in the figure. According to this diagram, which of the following option(s) is/are correct?
Show Hint
- The equilibrium inter-nuclear distance of Q is more than that of P
- The total energy \( E = 0 \) separates bound and unbound states of the molecules
- The lowest vibrational frequency of P is larger than that of Q
- Dissociation energy of Q is more than that of P
The Correct Option is A, B, C
Solution and Explanation
The question presents a potential energy diagram for two diatomic molecules, P and Q. Let's analyze each option based on the provided diagram:
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Equilibrium Inter-Nuclear Distance:
The equilibrium inter-nuclear distance corresponds to the minimum of the potential energy curve. From the diagram, the equilibrium inter-nuclear distance for molecule Q is larger than that for molecule P.
✓ Option (A) is correct.
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Energy E = 0 Separates Bound and Unbound States:
The energy level E = 0 represents the threshold between bound and unbound molecular states. States with energy below zero are bound (stable), and those above zero are unbound (dissociative).
✓ Option (B) is correct.
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Lowest Vibrational Frequency:
Vibrational frequency is linked to the curvature of the potential energy curve at the minimum. A steeper curve (as in P) indicates stronger restoring forces and thus a higher vibrational frequency compared to a shallower curve (as in Q).
✓ Option (C) is correct.
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Dissociation Energy:
Dissociation energy is the depth of the potential well. From the diagram, molecule Q has a deeper well, hence a higher dissociation energy than P.
✗ Option (D) is incorrect.
Conclusion: The correct options are (A), (B), and (C).
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