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JEE MainChemistryStructure of Atom

For a hydrogen atom, consider the following orbitals described by their quantum numbers: A. n=3, l=2, m_l=-2 B. n=3, l=2, m_l=+1 C. n=3, l=1, m_l=0 D. n=3, l=0, m_l=0 E. n=4, l=0, m_l=0 Which of the given orbitals will have the same energy?

Options

  1. AA, B, C and D only
  2. BA and B only
  3. CA, B and C only
  4. DD and E only

Correct answer

A. A, B, C and D only

Step-by-step solution

For a single-electron species like the hydrogen atom, the energy of an orbital depends strictly on the principal quantum number ( n ). The azimuthal quantum number ( l ) and magnetic quantum number ( m_l ) do not affect the energy. Analyzing the given options: A: n=3 B: n=3 C: n=3 D: n=3 E: n=4 Since orbitals A, B, C, and D all have the same principal quantum number ( n=3 ), they are degenerate and have the same energy. Orbital E has n=4 , so it has a different energy. Answer: A, B, C and D only

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