Step 1: Understand the relationship between the energies of lines.
The energy of a photon emitted in a transition between two levels in the hydrogen atom is given by the Rydberg formula:
\[
E = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)
\]
where:
- \( R_H \) is the Rydberg constant,
- \( n_1 \) and \( n_2 \) are the principal quantum numbers of the initial and final states.
Step 2: Compare the energies for different series.
- The Lyman series involves transitions to \( n_1 = 1 \) (from higher levels), and the energy of a photon increases as the transition is made from higher levels to \( n = 1 \).
- The Balmer series involves transitions to \( n_1 = 2 \) (from higher levels), so the energy of photons in the Balmer series is lower than that of the Lyman series for the same transition levels.
Step 3: Compare the lines within each series.
- In the Lyman series, the energy increases with the decreasing value of \( n_2 \). Thus, the energy of the second line of the Lyman series (D) will be greater than the first line (A).
- In the Balmer series, the energy decreases with increasing values of \( n_2 \). Thus, the energy of the first line of the Balmer series (C) will be greater than the second line (B).
Step 4: Order the energies.
- \( D \) (second line of Lyman series) has the highest energy.
- \( A \) (first line of Lyman series) is next in energy.
- \( B \) (second line of Balmer series) comes next.
- \( C \) (first line of Balmer series) has the lowest energy.
Thus, the correct decreasing order of energies is \( D>A>B>C \).