Discussion

In which state Be³⁺ will have same orbital radius as that of ground state of hydrogen atom ?

Answer: C

Be³⁺ has the hydrogen ground-state radius in the state n = 2.

For a hydrogen-like ion the Bohr radius of orbit n is

\[ r_{n} = \frac{n^{2}a_{0}}{Z} \]

For hydrogen in its ground state, n = 1 and Z = 1, so r = a₀. For Be³⁺, Z = 4, so \(r_{n} = a_{0}\) requires \(n^{2} = 4\), that is n = 2.

In n = 1 the Be³⁺ orbit is only a₀/4, and n = 3 gives 9a₀/4. n = 0 is not an allowed quantum number.

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