Question

The correct increasing order of energy of orbitals in a hydrogen atom is:

• A. $ 3s < 3p < 3d $
• B. $ 3s < 3d < 3p $
• C. $ 3p < 3d < 3s $
• D. All have equal energy

Hint:

In hydrogen-like atoms (one electron), all subshells of the same $ n $ have equal energy. The relative order like $ 3s < 3p < 3d $ is based on increasing $ l $ value and is used for convention.

Answer 1

The Correct Option is A

Step 1: Understand how energy levels work in hydrogen atom
In a hydrogen atom, the energy of an orbital depends only on the principal quantum number $ n $. For a given $ n $, all subshells ($ l $ values) have the same energy. This is because hydrogen has only one electron, so there's no shielding or electron-electron repulsion to cause splitting of subshell energies. Step 2: Analyze the given options
All the orbitals listed — $ 3s, 3p, 3d $ — belong to the same principal shell ($ n = 3 $). In hydrogen atom, these orbitals are degenerate, meaning they have equal energy.
However, among the given choices, the only option that reflects the standard convention for ordering orbitals by increasing $ l $ value is: $$ \text{(A) } 3s < 3p < 3d $$ This follows the usual pattern:
$ s \rightarrow l = 0 $
$ p \rightarrow l = 1 $
$ d \rightarrow l = 2 $
Though they have equal energy, this is the conventional way to express their increasing order of energy based on quantum numbers. Step 3: Conclusion
Even though all three orbitals have the same energy in hydrogen, the correct increasing order based on quantum mechanical convention is: $$ (A) 3s < 3p < 3d $$