Question

Which state of triply ionised Beryllium $(Be^{+++})$ has the same orbital radius as that of the ground state of hydrogen.
A.) n = 3
B.) n = 4
C.) n = 1
D.) n = 2

Answer 1

Hint: The beryllium is triply ionised, so it will have only one electron. So, we can use the radius of atom formula for hydrogen-like atoms to find that at what state of the triply ionised Beryllium, its radius is the same as that of ground state hydrogen.

Complete step by step solution
We have been given that the beryllium atom is triply ionised and its orbital radius is the same as that of the ground state of hydrogen.

Radius of an orbit of a hydrogen like atom can be found out by using the formula

${r}_{n}={r}_{0}\dfrac{n^2}{z}$, where ${r}_{n}$ is the radius of $n^{th}$ orbit of an atom, ${r}_{0}$ is the radius of hydrogen atom in its ground state, n is the number of orbit of the atom and z is the atomic number of the atom.

Here, z for beryllium is 4, and after three ionizations, its radius is equal to the radius of hydrogen atom in its ground state, that is ${r}_{n}={r}_{0}$. So, for that $\dfrac{n^2}{z}$ should be equal to one.

So, $\dfrac{n^2}{z}=1 \implies \dfrac{n^2}{4}=1 \implies n=2$.

Hence, option d is the correct answer.

Addition information:
A hydrogen-like atom basically comprises the species which when in ionized form have only one electron in its outermost orbit. Some examples of hydrogen-like atoms are $He^+$, $Li^{2+}$ and $Be^{3+}$.

Note: The formula is must in this question, so we should not forget this. One may also confuse the term ‘Hydrogen-like atom’ which is nothing but the atoms with one electron except that of hydrogen.