Question

If $$a_0$$ is the Bohr radius, the radius of then $$n=2$$ electronics orbit in triply ionized beryllium is:
A. $4a_0$
B. $$a_0$$
C. $$a_0/4$$
D. $$a_0/16$$

Answer 1

Hint: We know that atoms of different elements have different types of radius. There are many types of atomic radius such as a Van der Waals radius, Ionic radius, Covalent radius, Metallic radius and Bohr radius which explain the radius of atoms. We can solve this problem with the Bohr concept.

Complete step-by-step answer:
Atomic radius has a trend which varies with the increase in atomic number . In general , the atomic radius decreases as we move towards left in a period and have an increase in their radii when we move down a group. While the ionic radius follows similar trends as the atomic radius , ions may be larger or smaller than neutral atoms.
Bohr radius is the radius of the lowest-energy electron orbit of an atom. Bohr radius is only applicable to atoms and ions which have single electrons such as hydrogen, singly ionized helium etc. For calculating the radius of hydrogen equivalent system, we use following formula:
$r_n={\displaystyle \frac{n^{2}r}{Z}}$
Where, $Z$ is the atomic number of elements , $n$ is the number of orbits and $r$ is the total radius .
Applying the above formula in the given question we find that $r=a_0$and $n=2$. The atomic no. for beryllium is 4 so the value of $Z$ is equal to $4$.
$r_n={\displaystyle \frac{{\left(2\right)}^2{a}_0}{4}}$ $\Rightarrow a_0$
For the given question option B is correct.

So, the correct answer is “Option B”.

Note: According to Bohr concept, Bohr radius is the most probable distance between nucleus and the electron in a hydrogen atom in its ground states. The value of the Bohr radius is $5.29177\times {10}^{-11}m$.