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Orbital having 3 angular nodes and 3 total nodes is
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The correct answer is:
$4 f$
Angular node (l) $=3$
Total node $=$ radial node + angular node
$$
\begin{aligned}
& 3=(\mathrm{n}-\mathrm{l}-\mathrm{l})+\mathrm{l} \\
& 3=\mathrm{n}-\mathrm{l} \Rightarrow \mathrm{n}=4
\end{aligned}
$$
$\therefore$ Orbital having 3 angular nodes and 3 total nodes is $=n l=4 \mathrm{f}[\because \mathrm{l}=3$ for $\mathrm{f}$ - orbital $]$
Total node $=$ radial node + angular node
$$
\begin{aligned}
& 3=(\mathrm{n}-\mathrm{l}-\mathrm{l})+\mathrm{l} \\
& 3=\mathrm{n}-\mathrm{l} \Rightarrow \mathrm{n}=4
\end{aligned}
$$
$\therefore$ Orbital having 3 angular nodes and 3 total nodes is $=n l=4 \mathrm{f}[\because \mathrm{l}=3$ for $\mathrm{f}$ - orbital $]$
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