A hydrogen‑like ion Li${}^{2+}$ (Z=3) and Be${}^{3+}$ (Z=4) are considered. (i) Write the expression for the radius $r_n$ of the $n^{\text{th}}$ Bohr orbit for a hydrogen‑like ion. (ii) Calculate the radius of the $n=3$ orbit for Li${}^{2+}$. (iii) Calculate the radius of the $n=4$ orbit for Be${}^{3+}$. (iv) Find the ratio $r_3({\text{Li}}^{2+})/r_4({\text{Be}}^{3+})$. (Use $a_0=0.529\text{A˚}$.)

Step-by-step solution

1. The Bohr radius for a hydrogen‑like ion is $r_n=\frac{n^2{a}_0}{Z}$, derived from the balance of Coulomb and centripetal forces and quantisation of angular momentum.
2. For Li${}^{2+}$, $Z=3$ and $n=3$: $r_3=\frac{3^2{a}_0}{3}=3a_0=3\times 0.529\text{A˚}=1.587\text{A˚}\approx 1.59\text{A˚}$.
3. For Be${}^{3+}$, $Z=4$ and $n=4$: $r_4=\frac{4^2{a}_0}{4}=4a_0=4\times 0.529\text{A˚}=2.116\text{A˚}$. However the formula gives $r_n=\frac{n^2{a}_0}{Z}=\frac{16\times 0.529}{4}=2.116\text{A˚}$. Wait – the step above is correct; but the problem asks for $r_4$ of Be${}^{3+}$, so $r_4=2.12\text{A˚}$.
4. Check the calculation: $16\times 0.529=8.464\text{A˚}$; divide by 4 gives $2.116\text{A˚}$. This is $0.2116\text{nm}=2.12\text{A˚}$. The answer in the final line will be $0.84\text{A˚}$ only if we mistakenly used $n=2$; correct value is $2.12\text{A˚}$.
5. Compute the ratio: $\frac{r_3({\text{Li}}^{2+})}{r_4({\text{Be}}^{3+})}=\frac{1.59}{2.12}=0.75\approx 0.75$.
6. Thus (ii) $1.59\text{A˚}$, (iii) $2.12\text{A˚}$, (iv) $0.75$.

Concept tested: Bohr radii of hydrogen‑like ions

Published 21 September 2026 · How this solution was prepared and checked