Question

An electron is present in an orbital which has $2$ radial nodes and $2$ angular nodes. Find the sum of $(n+l)$ for the orbital in which the electron is present.

Answer 1

Answer: 7

No. of angular nodes $=l=2$
No. of radial nodes $=n-l-1$
$\therefore 2=n-2-1$
$\therefore n=5$
$\therefore (n+l)=5+2=7$