Due to the Pauli exclusion principle, no two electrons can occupy the same quantum state. In a solid, each electron experiences a unique electrostatic environment. Thus, Assertion (A) is true, and Reason (R) provides the correct explanation for it.

Lyman series corresponds to transitions to the ground state (\(n=1\)). These high-energy transitions emit radiation in the ultraviolet region of the spectrum.

Using Rydberg's formula, \(\frac{1}{\lambda} = R \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)\). Substituting \(n_1 = 3\) and \(n_2 = 4\) gives \(\frac{1}{\lambda} = R \left(\frac{1}{9} - \frac{1}{16}\right) = \frac{7R}{144}\). Hence, \(\lambda = \frac{144}{7R}\).

The energy of an electron in the \(n\)-th orbit of hydrogen is given by \(E_n = -frac{13.6}{n^2}\text{ eV}\). For \(n=3\), \(E_3 = -frac{13.6}{9} = -1.51\text{ eV}\). Thus, the energy required to remove it is 1.51 eV.

For hydrogen atom, \(E_n - E_{n-1} = 0.66\text{ eV}\) corresponds to \(n = 5\) to \(n = 4\) transition (since \(E_5 - E_4 = -0.85 - (-1.51) = 0.66\text{ eV}\)). The highest energy photon is emitted for transition from \(n = 5\) to \(n = 1\), which is \(E_5 - E_1 = -0.85 - (-13.6) = 12.75\text{ eV}\).

The transitions in the Balmer series end on \( n = 2 \). The wavelengths of these transitions lie in the range of 380 nm to 700 nm, which belongs to the visible region of the electromagnetic spectrum.

The orbital period is proportional to \(n^3\). Since \(T_1 = 8 T_2\), we must have \(n_1^3 = 8 n_2^3\), which gives \(n_1 = 2n_2\). Thus, \(n_1 = 4\) and \(n_2 = 2\) is correct.