$(A)$ $n=4,1=1$ $(B)$ $\mathrm{n}=4,1=2$ $(C)$ $\mathrm{n}=3,1=1$ $(D)$ $\mathrm{n}=3,1=2$ $(E)$ $n=4,1=0$
નીચે આપેલા વિકલ્પોમાંથી સાચો જવાબ પસંદ કરો.
\((A)\) \(\mathrm{n}=4, \ell=1 \Rightarrow(\mathrm{n}+\ell)=5\)
\((B)\) \(\mathrm{n}=4, \ell=2 \Rightarrow(\mathrm{n}+\ell)=6\)
\((C)\) \(\mathrm{n}=3, \ell=1 \Rightarrow(\mathrm{n}+\ell)=4\)
\((D\)) \(\mathrm{n}=3, \ell=2 \Rightarrow(\mathrm{n}+\ell)=5\)
\((E)\) \(\mathrm{n}=4, \ell=0 \Rightarrow(\mathrm{n}+\ell)=4\)
For same value of \((n+\ell)\), orbital having higher value of \(n\), will have more energy.
\((B) \)>\( (A) \)>\( (D) \)>\( (E) \)>\( (C)\)
$n$ $l$ $m$