1. Given state of excitation

   The hydrogen atom is excited from the ground state to the state with principal quantum number $n=4.$

2. Possible downward transitions

   When the electron returns from $n=4$ to lower levels, spectral lines are produced for every possible transition between the available levels: $n=4,3,2,1.$

   The possible emission transitions are:

   • $4 \to 3$
   • $4 \to 2$
   • $4 \to 1$
   • $3 \to 2$
   • $3 \to 1$
   • $2 \to 1$

   So total number of distinct spectral lines $= 6$.

3. Formula method

   If an atom is excited to level $n$, then the maximum number of spectral lines emitted is $N=\frac{n(n-1)}{2}.$

   For $n=4$, $N=\frac{4(4-1)}{2}=\frac{12}{2}=6.$

4. Option check

   • A: $2$ ✗
   • B: $3$ ✗
   • C: $5$ ✗
   • D: $6$ ✓

Therefore, the correct answer is D.