The correct decreasing order of energy for the orbitals having, following set of quantum numbers :
(A) n = 3, l = 0, m = 0
(B) n = 4, l = 0, m = 0
(C) n = 3, l = 1, m = 0
(D) n = 3, l = 2, m = 1
is :
Solution
(A) $\mathrm{n}+\ell=3+0=3$<br/><br/>
(B) $\mathrm{n}+\ell=4+0=4$<br/><br/>
(C) $\mathrm{n}+\ell=3+1=4$<br/><br/>
(D) $\mathrm{n}+\ell=3+2=5$<br/><br/>
Higher $\mathrm{n}+\ell$ valuc, higher the encrgy & if same $\mathrm{n}+\ell$ value, then higher $\mathrm{n}$ value, higher the energy.<br/><br/>
Thus : $\mathrm{D}>\mathrm{B}>\mathrm{C}>\mathrm{A}$.
About this question
Subject: Chemistry · Chapter: Atomic Structure · Topic: Bohr's Model
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