Question

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

(a) hydrogen
(b) deuterium
(c) He⁺
(d) Li⁺⁺

Answer 1

(d) Li⁺⁺

The radius of the nth orbit in one electron system is given by
$r_{\mathrm{n}}=\frac{n^2{a}_0}{Z}$

Here, a₀ = 53 pm

For the shortest orbit,
n = 1

For hydrogen,
Z = 1

∴ Radius of the first state of hydrogen atom = 53 pm


For deuterium,
Z= 1

∴ Radius of the first state of deuterium atom = 53 pm


For He⁺,
Z = 2

∴ Radius of He⁺ atom = $\frac{53}{2}\mathrm{pm}=26.5\mathrm{pm}$


For Li⁺⁺,
Z = 3

∴ Radius of Li⁺⁺ atom = $\frac{53}{3}\mathrm{pm}=17.66\mathrm{pm}\approx 18\mathrm{pm}$

The given one-electron system having radius of the shortest orbit to be 18 pm may be Li⁺⁺.