For `He^(+)` the electron is in orbit with energy equal to 3.4 eV. The aximuthal quantum numer for that orbit is 2 and magnetic quantum numbr is 0. Then which of the following is correct

Options

A. The subshell is 5d.
B. The number of angular nodes in it is 2.
C. The number of radial nodes in it is 3.
D. The nuclear charge experienced in n=4 is 2 e less than that in n=1 where e is electronic charge

Answer

B

Explanation

`underset(n^(2)=16)(E_(He^(+))=-)13.6xxunderset(so)(((2)^(2))/(n^(2)))=-underset(n=4)(3.4=)(-13.6)/(4)`
Quantum number are
`n=4,l=2,m=0`
so subshell m=0
angular node =l=2
radial node `=[n-l-1]=4-2-1=1`