If [ 0 . 53 × 10 ^ - 10 4 ] metre is the radius of smallest...

For hydrogen like atom, the radius of n^{th} orbit is given by,

${r}_{n} = \dfrac{n2}{Z}{r}_{0}$

Where, ${r}_{0} = 0. 53 \times {10}^{- 10}m$

and ${r}_{n} = \dfrac{0. 53 \times {10}^{- 10}m}{4}$

At ground state, $n = 1$

Hence, the atom is ${Be}^{3 +} [ ∵ Z = 4 ]$

So, the atom is triply ionised beryllium ${Be}^{3 +}$.