Question

A hydrogen atom and $L{i}^{++}$ ion are both in the second excited state. If $I_H$ and $I_{Li}$ are their respective electronic angular momenta, and $E_H$ and $E_L$i are their respective energies, then

• A. $I_H>I_{Li}$ and $|E_H|>|E_{Li}|$
• B. $I_H=I_{Li}$ and $|E_H|<|E_{Li}|$
• C. $I_H=I_{Li}$ and $|E_H|>|E_{Li}|$
• D. $I_H<I_{Li}$ and $|E_H|<|E_{Li}|$

Answer 1

The correct option is B $I_H=I_{Li}$ and $|E_H|<|E_{Li}|$

Given $Z_H=1,Z_{Li}=3$ and $n_H=n_{Li}=3$
1. Angular momentum is independent of Z and proportional to $n$ as $mvr=\frac{nh}{2\pi }$, therefore $I_H=I_{Li}$

2. $E=-13.6\frac{Z^2}{n^2}$
Energy is directly proportional to square of atomic number, hence $E_H<E_{Li}$