The correct decreasing order of energy, for the orbitals having, following set of quantum numbers:
(A) $\mathrm{n}=3, \ell=0, \mathrm{~m}=0$
(B) $n=4, \ell=0, m=0$
(C) $\mathrm{n}=3, \ell=1, \mathrm{~m}=0$
(D) $n=3, \ell=2, m=1$
Select the correct option:
A
(D) > (B) > (C) > (A)
B
(B) > (D) > (C) > (A)
C
(C) > (B) > (D) > (A)
D
(B) > (C) > (D) > (A)
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
(A) $n+\ell=3+0=3$
(B) $n+\ell=4+0=4$
(C) $n+\ell=3+1=4$
(D) $n+\ell=3+2=5$
Higher $\mathrm{n}+\ell$ value, higher the energy and if same $\mathrm{n}+\ell$ value, then higher n value, higher the energy.
Thus: $\mathrm{D}>\mathrm{B}>\mathrm{C}>\mathrm{A}$.
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