Question

(a) Derive a general expression for the energy difference between any two neighboring rotational state (i.e, the difference in energy between the L abd L+L states). (b) How does the energy difference change as L increases? (c) How does the energy difference change if the radius, r, of the rotating particle increases?

Answer 1

a) The energy difference between any two neighbouring rotational states can be derived using Law of conservation of energy, which in this case is written as Mechanical energy = Rotational Kinetic energy+ Potential energy ± forces between two particles (which can be attractive or repulsive)

m₁ & m₂ are masses of particles rotating in states L and L+L respectively, 'r' is the radius between the two rotating particles which in turn depends upon their postions.

ω₁& ω₂ are angular velocities

Writing energy equation for L state,

E₁= 1/2(Σm_ir_i²)_Lω₁² + m₁g(0) ± Gm₁m₂/r²   

( Σm_ir_i²can also be written as I(inertia of motion), as it is the sum of individual kinetic energies of each of the sub-particles on a particle)

Similarly we can write energy equation for L+L State,

E₂=1/2(Σm_ir_i²)_L+Lω₂² + m₂g(L) ± Gm₁m₂/r²   ( distance between 2 states as ' L')

therefore difference between two energy states will be,

E₂ -E₁= [1/2(Σm_ir_i²)_Lω₁² + m₁g(0) ± Gm₁m₂/r² ] - [1/2(Σm_ir_i²)_L+Lω₂² + m₂g(L) ± Gm₁m₂/r² ]

So, the final equation would be

ΔE = E₂ -E₁= 1/2 { (Σm_ir_i²)_Lω₁² - (Σm_ir_i²)_L+L ω₂² } + m₂g(L) ± 2Gm₁m₂/r² .

(b) from the above equation it is clear that ΔE is directly proportional to the L (ΔE∝  m₂g(L) ), therefore the energy difference increases as L increases.

(c) This can also be known from the derived final expression, as  ΔE ∝ GGm₁m₂/r² (ΔE ∝1/r², inversely proportional ), hence the energy difference decreases gradually as radius 'r' of the the rotating particle increases.