Question

Radiation coming from transitions $n=2$ to $n=1$ of hydrogen atoms fall on $H{{e}^{+}}$ ions in $n=1$ and $n=2$ states. When they absorb energy from the radiation, the possible transition of helium ions will be,
$\begin{align}
  & A.n=1\to n=4 \\
 & B.n=2\to n=4 \\
 & C.n=2\to n=5 \\
 & D.n=2\to n=3 \\
\end{align}$

Answer 1

Hint: Energy released by the helium atom can be found by taking the product of the energy of the first orbit of the hydrogen atom, the square of the atomic number and the difference between the reciprocal of the square of the number of shells. First of all find out the energy released due to the hydrogen atom and then compare it with that of the helium atom. This will help you in answering this question.

Complete step by step answer:
Energy released during the transition from $n=2$ to $n=1$ of the hydrogen atom. The energy released can be written as an equation,
$E=13.6{{Z}^{2}}\left( \dfrac{1}{{{n}_{1}}^{2}}-\dfrac{1}{{{n}_{2}}^{2}} \right)$
As per the question, for the hydrogen atom,
$Z=1,{{n}_{1}}=1,{{n}_{2}}=2$
Substitute this in the equation will give,
\[E=13.6\times 1\left( \dfrac{1}{{{1}^{2}}}-\dfrac{1}{{{2}^{2}}} \right)=13.6\times \dfrac{3}{4}eV\]
Now we have to compare this value with the energy released on each of the transitions mentioned in the question. The one with the value equivalent to that of a hydrogen atom will be the answer for the question.
As we all know, the atomic number of this ion will be,
\[Z=2\]
The energy released due to $H{{e}^{+}}$in the first case will be,
\[n=1\to n=4\]
Using this values in the equation will give,
\[E=13.6\times {{2}^{2}}\left( \dfrac{1}{{{1}^{2}}}-\dfrac{1}{{{4}^{2}}} \right)=13.6\times \dfrac{15}{4}eV\]
This is not equal to that of the hydrogen atom. Therefore this is not the right answer.
In second case,
\[n=2\to n=4\]
The energy released will be,
\[E=13.6\times {{2}^{2}}\left( \dfrac{1}{{{2}^{2}}}-\dfrac{1}{{{4}^{2}}} \right)=13.6\times \dfrac{3}{4}eV\]
This will be a possibility to be the answer. Let us have all other options too.
In third case,
\[n=2\to n=5\]
The energy released will be,
\[E=13.6\times {{2}^{2}}\left( \dfrac{1}{{{2}^{2}}}-\dfrac{1}{{{5}^{2}}} \right)=13.6\times \dfrac{21}{25}eV\]
And for the fourth case,
\[n=2\to n=3\]
The energy released will be,
\[E=13.6\times {{2}^{2}}\left( \dfrac{1}{{{2}^{2}}}-\dfrac{1}{{{3}^{2}}} \right)=13.6\times \dfrac{5}{9}eV\]
Therefore, Energy needed for transition of $H{{e}^{+}}$ for \[n=2\to n=4\] matches exactly with energy released in transition of \[H\] for \[n=2\to n=1\].
The answer will be option B.

Note:
Molecular electronic transitions occur if the electrons in a molecule are excited from one energy level to a greater energy level. This energy difference associated with this transition will give the idea about the structure of a molecule and detect many molecular characteristics.