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
- AA, B, C and D only
- BA and B only
- CA, B and C only
- 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