C

According to Bohr's model, the time period (T) of an electron in an orbit is given by:

$$ T_n = \frac{2\pi}{v_n} = 2\pi \sqrt{\frac{m r_n^3}{k_e e^2}} $$
and the radius (r) of the orbit is given by:
$$ r_n = n^2 \frac{h^2}{4\pi^2 m e^2} $$
Therefore, we can express the time period as:
$$ T_n \propto n^3 $$
If we assume the relation between the time periods in two different states is:
$$ \frac{T_{n_1}}{T_{n_2}} = \left( \frac{n_1}{n_2} \right)^3 $$
Given that the time period of the electron in the initial state (n_1) is eight times that in the final state (n_2), we have:
$$ T_{n_1} = 8 T_{n_2} $$
Substituting into our relation yields:
$$ \frac{T_{n_1}}{T_{n_2}} = 8 = \left( \frac{n_1}{n_2} \right)^3 $$
Taking the cube root gives:
$$ \frac{n_1}{n_2} = 2 $$
Therefore, if we let n_2 = n, we have n_1 = 2n.
The possible integer values for n_1 and n_2 that satisfy the principal quantum number conditions are:
- If n_2 = 1, then n_1 = 2 * 1 = 2
- If n_2 = 2, then n_1 = 2 * 2 = 4
Hence, the possible pairs (n_1, n_2) are (2, 1) and (4, 2).

The only valid transition allowed under quantum mechanics for hydrogen is (2, 1). Therefore, options C = (2, 1) is the only valid transition.