Compare the energies of following sets of quantum numbers for multielectron system.
(A) $\mathrm{n}=4,1=1$ (B) $\mathrm{n}=4,1=2$
(C) $n=3,1=1$ (D) $\mathrm{n}=3,1=2$
(E) $n=4,1=0$
Choose the correct answer from the options given below:
Select the correct option:
A
(B) > (A) > (C) > (E) > (D)
B
(E) > (C) < (D) < (A) < (B)
C
(E) > (C) > (A) > (D) > (B)
D
(C) < (E) < (D) < (A) < (B)
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
Energy level can be determined by comparing ( $n+\ell$ ) values
(A) $\mathrm{n}=4, \ell=1 \Rightarrow(\mathrm{n}+\ell)=5$
(B) $\mathrm{n}=4, \ell=2 \Rightarrow(\mathrm{n}+\ell)=6$
(C) $n=3, \ell=1 \Rightarrow(n+\ell)=4$
(D) $\mathrm{n}=3, \ell=2 \Rightarrow(\mathrm{n}+\ell)=5$
(E) $\mathrm{n}=4, \ell=0 \Rightarrow(\mathrm{n}+\ell)=4$
For same value of ( $n+\ell$ ), orbital having higher value of $n$, will have more energy.
$(\mathrm{B})>(\mathrm{A})>(\mathrm{D})>(\mathrm{E})>(\mathrm{C})$
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