The region in the electromagnetic spectrum where the Balmer series lines appear is:

We begin by recalling the general formula for the wavelength of any spectral line in the hydrogen atom. For a transition that ends in an energy level of principal quantum number $$n_l$$ and starts from a higher level $$n_h$$ (where $$n_h > n_l$$), the well-known Balmer-Rydberg formula is stated as

$$\frac{1}{\lambda}=R\left(\frac{1}{n_l^{\,2}}-\frac{1}{n_h^{\,2}}\right),$$

where $$\lambda$$ is the wavelength of the emitted photon and $$R$$ is the Rydberg constant $$R\approx1.097\times10^{7}\,\text{m}^{-1}.$$

Balmer himself chose $$n_l = 2$$, so every line of the Balmer series corresponds to electrons falling to the second orbit. Therefore, for the Balmer series we have

$$\frac{1}{\lambda}=R\left(\frac{1}{2^{2}}-\frac{1}{n_h^{\,2}}\right)=R\left(\frac{1}{4}-\frac{1}{n_h^{\,2}}\right),\qquad n_h = 3,4,5,\ldots$$

Now let us see what orders of magnitude for $$\lambda$$ this formula yields. We may take a few values of $$n_h$$ just to convince ourselves numerically:

For $$n_h = 3$$ (the first Balmer line, called Hα):

$$\frac{1}{\lambda}=R\left(\frac{1}{4}-\frac{1}{9}\right)=R\left(\frac{9-4}{36}\right)=R\left(\frac{5}{36}\right).$$

Substituting $$R=1.097\times10^{7}\,\text{m}^{-1}$$, we get

$$\frac{1}{\lambda}=1.097\times10^{7}\times\frac{5}{36}=1.523\times10^{6}\,\text{m}^{-1},$$

so

$$\lambda=\frac{1}{1.523\times10^{6}}=6.57\times10^{-7}\,\text{m}=657\,\text{nm}.$$

This is clearly inside the range $$400 \text{ nm} \le \lambda \le 700 \text{ nm}$$, which we recognize as the visible region.

For $$n_h = 4$$ (Hβ):

$$\frac{1}{\lambda}=R\left(\frac{1}{4}-\frac{1}{16}\right)=R\left(\frac{4-1}{16}\right)=R\left(\frac{3}{16}\right).$$

So

$$\frac{1}{\lambda}=1.097\times10^{7}\times\frac{3}{16}=2.055\times10^{6}\,\text{m}^{-1},$$

yielding

$$\lambda=\frac{1}{2.055\times10^{6}}=4.87\times10^{-7}\,\text{m}=486\,\text{nm},$$

again comfortably in the visible band.

Proceeding to still higher $$n_h$$ values only pushes the wavelength a little further toward the violet end but always within or very near the visible limits before merging into the ultraviolet convergence limit at $$\lambda\approx364\,\text{nm}$$. The main intense Balmer lines lie wholly in the visible window.

Because the Balmer series wavelengths fall between roughly $$364\,\text{nm}$$ and $$656\,\text{nm}$$, the region of the electromagnetic spectrum in which these lines are observed is indeed the visible region.

Hence, the correct answer is Option A.