Question

The radius of shortest orbit in one electron system is 18 pm. It may be
(A) ${}_1^1H$
(B) ${}_2^1H$
(C) $H{e^ + }$
(D) $L{i^ + }$

Answer 1

Hint
Here we use the formula which is used to find the radius of any atom or an electron system. It is given by,
${r_n} = \dfrac{{{n^2}{a_0}}}{Z}$
Where $r_n$ is the radius of the system, $n$ implies the $n^{th}$ orbit of the electron, $r_B$ is the Bohr’s radius ({a_0} = 0.53\mathop A\limits^0 = 53pm) and Z is the charge of the nucleus.

Complete step by step solution
Given, the radius of the shortest orbit in one electron system, $r_n = 18$ pm.
Since a shortest orbit is to be considered, we should take the least value of $n$. The least value $n$ can take is $1$. So, $n = 1$.
Now, let us find the radius of each of the given electron systems.
For ${}_1^1H, Z = 1; \therefore {r_n} = \dfrac{{{1^2} \times 53pm}}{1} = 53$pm.
But, the given radius is 18 pm. So, option (A) is incorrect.
For ${}_2^1H, Z = 1; \therefore {r_n} = \dfrac{{{1^2} \times 53pm}}{1} = 53$pm
But, the given radius is 18 pm. So, option B is incorrect.
For $H{e^ + }, Z = 2; \therefore {r_n} = \dfrac{{{1^2} \times 53pm}}{2} = 26.5$pm
But, the given radius is 18 pm. So, option C is incorrect.
For $L{i^ + }, Z = 3; \therefore {r_n} = \dfrac{{{1^2} \times 53pm}}{3} = 17.667pm = 18$pm
This is the given radius. Therefore, the one electron system with radius 18 pm is $Li^+$.
The correct answer is option (D).

Note
There is an alternative method to solve such kinds of questions. Consider the radius formula given by, ${r_n} = \dfrac{{{n^2}{a_0}}}{Z}$, where $r_n$ is the radius of the system, $n$ implies the $n^{th}$ orbit of the electron, $r_B$ is the Bohr’s radius $({a_0} = 0.53\mathop A\limits^0 = 53pm)$ and $Z$ is the charge of the nucleus.
Now, rearrange the radius formula in terms of $Z$.
$Z = \dfrac{{{n^2}{r_B}}}{{{r_n}}}$
Substitute the given values and calculate the value of $Z$.
$\Rightarrow Z = \dfrac{{{1^2} \times 0.53\mathop A\limits^0 }}{{0.18\mathop A\limits^0 }} \Rightarrow Z = 2.944 \therefore Z = 3$
Among the given one-electron systems, the system with $Z = 3$.