MCQ
The hydrogen-like species $\mathrm{Li}^{2+}$ is in a spherically symmetric state $\mathrm{S}_1$ with one radial node. Upon absorbing light the ion undergoes transition to a state $\mathrm{S}_2$. The state $\mathrm{S}_2$ has one radial node and its energy is equal to the ground state energy of the hydrogen atom.

$1.$ The state $S_1$ is

$(A)$ $1 \mathrm{~s}$ $(B)$ $2 \mathrm{~s}$ $(C)$ $2 \mathrm{p}$ $(D)$ $3 \mathrm{~s}$

$2.$ Energy of the state $S_1$ in units of the hydrogen atom ground state energy is

$(A)$ $0.75$ $(B)$ $1.50$ $(C)$ $2.25$ $(D)$ $4.50$

$3.$ The orbital angular momentum quantum number of the state $S_2$ is

$(A)$ $0$ $(B)$ $1$ $(C)$ $2$ $(D)$ $3$

Give the answer question $1,2$ and $3.$

  • A
    $(A,B,C)$
  • $(B,C,B)$
  • C
    $(A,B,A)$
  • D
    $(D,B,C)$

Answer

Correct option: B.
$(B,C,B)$
b
$1.$ For, $\mathrm{S}_1$ (spherically symmetrical)

$ \text { node }=1 $

$ \Rightarrow \mathrm{n}-1=1 $

$ \mathrm{n}=2$

For $S_2$, radial node $=1$

$ E_{\mathrm{S}_2}=\frac{-13.6 \times z^2}{n^2}=E_H \text { in ground state }=-13.6 $

$ E=\frac{-13.6 \times 9}{n^2} \Rightarrow n=3$

So, state $S_1$ is $2 \mathrm{~s}$ and $S_2$ is $3 \mathrm{p}$.

$2.$ $\frac{E_{S_1}}{E_{\mathrm{H}(\text { ground })}}=\frac{-13.6 \times 9}{4 \times(-13.6)}=2.25$

$3.$ Azimuthal quantum number for $\mathrm{S}_2=\ell=1$

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