Question

The radius of the shortest orbit in a one electron system is 18pm. It may be

• A.
• B.
• C.
• D.

Answer 1

Let's list the atomic numbers, the values for the given atoms –
$\begin{array}{cc}Element& Atomicno.,Z\\ Hydrogen,H& 1\\ Deuterium\left(ahydrogenisotope\right),D& 1\\ Heliumion,H{e}^{+}& 2\\ Lithiumion,L{i}^{++}& 3\end{array}$
We have learnt that the atomic radius really depends upon two numbers: the principal quantum number n, and the atomic number of the atom, Z. Since we are talking about the radius of the shortest orbit, it must be for n=1, and the given value is r=18 pm. Following the relation –
$r=\frac{n^2{a}^0}{Z}$,
Where $a_0=53$ pm is the shortest orbital radius for hydrogen, also called the Bohr radius, if we put n=1, we have –
$r=\frac{a_0}{Z}=\frac{53}{Z}pm=18pm$
$\Rightarrow Z=\frac{53}{18}\approx 3.$