Consider a particle on a ring that is perturbed by interacting with an applied electric field (E) with the perturbation being H' = μE cos Φ, where μ is the dipole moment. The energy levels correct upto first order are

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CSIR-UGC (NET) Chemical Science: Held on (26 Nov 2020)
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  1. \(\frac{m^2_ℓh^2}{2}-\frac{\mu E}{\pi}\)
  2. \(\frac{m^2_ℓh^2}{2I}-\frac{\mu E}{2\pi}\)
  3. \(\frac{m^2_ℓh^2}{2I}+\frac{\mu E}{2\pi}\)
  4. \(\frac{m^2_ℓh^2}{2I}\)

Answer (Detailed Solution Below)

Option 4 : \(\frac{m^2_ℓh^2}{2I}\)
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Detailed Solution

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Concept:

  • Perturbation theory is an approximation method to find out the exact solution of any system with great accuracy.
  • To break the energy correction in a particular energy level, we have to disturb the operator in a specific order.
  • The energy correction is given by,

\(\Delta E = \int {{\Psi ^ * }_ \circ {H^I}} {\Psi _ \circ }d\tau \)

  • The perturbed energy levels upto first order are given by,

\(E = {E^{\left( 0 \right)}} + {E^{\left( I \right)}}\)

  • For a particle in a ring, the ground state quantized energy values are,

\({E^{\left( 0 \right)}} = {{{m_l}^2{\hbar ^2}} \over {2l}}\), where ml is the magnetic quantum number.

Explanation:

  • For a particle in a ring that is perturbed by interacting with an applied electric field (E) with the perturbation being

\({H^I} = \mu E\cos \phi \)

  • For a particle in a ring, the wavefunction is given by,

​ \(\Psi = N{e^{in\phi }}\)

  • The first-order perturbed energy (\({E^{\left( I \right)}}\)) is given by,

\({E^{\left( I \right)}} = \int_0^{2\pi } {{\Psi _ \circ }\left( {\mu E\cos \phi } \right){\Psi _ \circ }d\phi } \)

\( = \int_0^{2\pi } {N{e^{in\phi }}\left( {\mu E\cos \phi } \right)N{e^{ - in\phi }}d\varphi } \)

\( = {N^2}\mu E\int_0^{2\pi } {{e^{in\phi - in\phi }}\cos \phi d\phi } \)

\( = {N^2}\mu E\int_0^{2\pi } {\cos \phi d\phi } \)

\( = {N^2}\mu E[\sin 2\pi - \sin 0]\)

=0

Conclusion:

  • Hence, energy levels correct upto first order are,

\(E = {E^{\left( 0 \right)}} + {E^{\left( I \right)}}\)

\( = {{{m_l}^2{\hbar ^2}} \over {2l}} + 0\)

\( = {{{m_l}^2{\hbar ^2}} \over {2l}}\)

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