The electron in the hydrogen atom undergoes transition from higher orbitals to orbital of radius 211.6 pm. This transition is associated with

For any hydrogenic atom, Bohr’s model gives the radius of the $$n^{\text{th}}$$ orbit as

$$r_n = n^2 a_0,$$

where $$a_0$$ is the Bohr radius. Numerically,

$$a_0 = 0.529\;\text{Å} = 0.529 \times 10^{-10}\;\text{m} = 52.9\;\text{pm}.$$

We are told that after the electron finishes its transition, it is found in an orbit whose radius is

$$r_n = 211.6\;\text{pm}.$$

Using the formula $$r_n = n^2 a_0$$, we can solve for $$n$$:

$$n^2 = \dfrac{r_n}{a_0}.$$

Substituting the numerical values, we have

$$n^2 = \dfrac{211.6\;\text{pm}}{52.9\;\text{pm}}.$$

Carrying out the division,

$$n^2 = \dfrac{211.6}{52.9} = 4.00.$$

So,

$$n = \sqrt{4.00} = 2.$$

This means the electron finally lands in the second orbit (the $$n = 2$$ level) after completing its transition.

In the hydrogen spectrum, all transitions that terminate at $$n = 2$$ constitute the Balmer series. Therefore, the radiation observed from this transition belongs to the Balmer region.

Hence, the correct answer is Option B.