Final answer:
The maximum number of spectral lines emitted when an electron in the third excited state (n=4) of a hydrogen atom transitions to lower energy states is six. This is calculated using the formula N = n(n - 1)/2.
Step-by-step explanation:
It refers to an electron in the third excited state in a hydrogen atom undergoing transitions to lower energy states and asks for the maximum number of spectral lines that can be emitted during these transitions.
In Bohr's model of the hydrogen atom, an excited state corresponds to the electron being in an orbit further from the nucleus than the ground state. Since the third excited state of hydrogen is represented by n=4 (where n is the principal quantum number), the transitions can occur to the n=3, n=2, and n=1 energy levels.
The maximum number of spectral lines that can be emitted is given by the formula for the number of transitions, N = n(n - 1)/2, where n is the principal quantum number of the initial excited state. For the third excited state, n=4, so N = 4(4 - 1)/2, which gives us 6. This means the maximum number of spectral lines that can be emitted is six.