Since comments caused certain level of confusion, I guess I'll try to provide a further illustration. You should consider all possibilities for an electron "jumping" down the excited energy state $n$ to the ground state $n = 1$. Electron doesn't get stuck forever on any of the levels with $n > 1$. Besides that, spectra is not a characteristic of a single excited atom, but an ensemble of many and many excited hydrogen atoms. In some atoms electrons jump directly from $n = 6$ to $n = 1$, whereas in some others electrons undergo a cascade of quantized steps of energy loss, say, $6 → 5 → 1$ or $6 → 4 → 2 → 1$. The goal is to achieve the low energy state, but there is a finite number of ways $N$ of doing this. I put together a rough drawing in Inkscape to illustrate all possible transitions*: I suppose it's clear now that each energy level $E_i$ is responsible for $n_i - 1$ transitions (try counting the colored dots). To determine $N$, you need to sum the states, as Soumik Das rightfully commented: $$N = \sum_{i = 1}^{n}(n_i - 1) = n - 1 + n - 2 + \ldots + 1 + 0 = \frac{n(n-1)}{2}$$ For $n = 6$: $$N = \frac{6(6-1)}{2} = 15$$ Obviously the same result is obtained by taking the sum directly. * Not to scale; colors don't correspond to either emission spectra wavelenghts or spectral series and solely used for distinction between electron cascades used for the derivation of the formula for $N$. Answer VerifiedHint:Electrons get excited when it absorbs energy. When electrons are excited, it jumps from ground state to excited state. The ground state has a lower energy while the excited state has a higher energy. The energy is released when the electrons return back to the ground state. Complete answer: When the electron transitions occur in atoms or ions, spectral lines are formed. These spectral lines have different wavelengths. There are six series by which electron transition occurs. They are Lymann, Balmer, Paschen, Brackett, Pfund and Humphrey series. When electrons get excited, photons are emitted. The energy of photon is given below:${{h}}\upsilon = {{{E}}_{{f}}} - {{{E}}_{{i}}}$, where ${{h}}$ is the Planck’s constant, $\upsilon $ is the frequency, ${{{E}}_{{f}}}$ is the final energy state and ${{{E}}_{{i}}}$ is the initial energy state. It is given that the electron of a hydrogen atom jumps from ${{n = 4}}$ to ${{n = 1}}$ state. Based on Bohr model of hydrogen spectrum, the number of spectral lines emitted can be calculated by the formula given below:\[ = \dfrac{{\left( {{{{n}}_2} - {{{n}}_1}} \right)\left( {{{{n}}_2} - {{{n}}_1} - 1} \right)}}{2} = \dfrac{{\left( {4 - 1} \right)\left( {4 - 1 + 1} \right)}}{2} = \dfrac{{3 \times 4}}{2} = 6\]When the electron of a hydrogen atom jumps from ${{n = 4}}$ to ${{n = 1}}$ state, the number of spectral lines will be $6$.Hence, the correct option is B. Note: Different series of spectral lines have different uses. For example, the Balmer series has been used in the astronomy field since it appeared in stellar objects. Sometimes, in the question, ${{{n}}_1}$ may not be given. In this case, another formula can be used. The formula is $\dfrac{{{{n}}\left( {{{n}} - 1} \right)}}{2}$, where ${{n}}$ is the higher energy level. Sometimes, the electron comes back from excited state to ground state directly. Sometimes it may take several steps. But the energy released is the same as that the energy absorbed.Open in App Suggest Corrections 8 |