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:
-
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.
-
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.
-
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.
-
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).
Questions Asked in GATE PH exam
A wheel of mass \( 4M \) and radius \( R \) is made of a thin uniform distribution of mass \( 3M \) at the rim and a point mass \( M \) at the center. The spokes of the wheel are massless. The center of mass of the wheel is connected to a horizontal massless rod of length \( 2R \), with one end fixed at \( O \), as shown in the figure. The wheel rolls without slipping on horizontal ground with angular speed \( \Omega \). If \( \vec{L} \) is the total angular momentum of the wheel about \( O \), then the magnitude \( \left| \frac{d\vec{L}}{dt} \right| = N(MR^2 \Omega^2) \). The value of \( N \) (in integer) is:
- GATE PH - 2025
- Rotational Motion
- The Hamiltonian for a one-dimensional system with mass \( m \), position \( q \), and momentum \( p \) is: \[ H(p, q) = \frac{p^2}{2m} + q^2 A(q) \] where \( A(q) \) is a real function of \( q \). If \[ m \frac{d^2 q}{dt^2} = -5q A(q), \] then \[ \frac{d A(q)}{d q} = n \frac{A(q)}{q}. \] The value of \( n \) (in integer) is:
- GATE PH - 2025
- Mechanics
- A system of three non-identical spin-\(\frac{1}{2}\) particles has the Hamiltonian
\[ H = \frac{A \hbar^2}{2} (\vec{S}_1 + \vec{S}_2) \cdot \vec{S}_3, \] where \( \vec{S}_1, \vec{S}_2, \vec{S}_3 \) are the spin operators of particles labelled 1, 2, and 3 respectively, and \( A \) is a constant with appropriate dimensions. The set of possible energy eigenvalues of the system is:- GATE PH - 2025
- Quantum Mechanics
- Nuclear radiation emitted from a \( {Co}^{60} \) radioactive source is detected by a photomultiplier tube (PMT) coupled to a scintillator crystal. Which of the following option(s) is/are correct?
- GATE PH - 2025
- Radiation Detection
- Which of the following consideration(s) is/are showing that nuclear beta decay, \( n \to p + e^{-} + \bar{\nu}_e \), has to be a three-body decay?
- GATE PH - 2025
- Nuclear Physics