Question

Electron of ${Be}^{3+}$ ion is present in an orbital which has 2 angular nodes and 2 radial nodes. Give $(n+\ell )$ for orbital in which electron is present.
(Given : In this question n is principles quantum number and $\ell$ is azimuthal quantum number)

Answer 1

No. of angular nodes = $l=2$

No. of radius node = $n-l-1=2$

$=n=5$

$\therefore n+l=7$