Maximum number of spectral lines emitted when an electron jumps from n=5 to n=1 in hydrogen atom is

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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 $N$.

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Hint :In order to solve this question we must know few basic information like what is meant by spectral lines. Spectral series are said to be the set of wavelengths arranged in a sequential fashion which characterizes light or any electromagnetic radiation emitted by the energised atoms.

Complete Step By Step Answer:

Hydrogen atom is said to be the simplest atomic system found in nature, thus it produces the simplest of the spectral series.When electrons de-excite from higher energy level ( $ n_2 $ ) to lower energy level ( $ n_1 $ ) in atomic sample, then number of spectral line observed in the spectrum is given by the formula; $ Total{\text{ }}no{\text{ }}of{\text{ }}the{\text{ }}spectral{\text{ }}lines\; = \dfrac{{\left( {n_2 - n_1} \right)\left( {n_2 - n_1 + 1} \right)}}{2} $ Since it is given in the question that in hydrogen atom electron jumps from $ n_2 = 5\; $ to $ n_1 = 1\; $ Therefore the total no of the spectral line is found to be; $ Total{\text{ }}no{\text{ }}of{\text{ }}the{\text{ }}spectral{\text{ }}lines\; = \dfrac{{\left( {5 - 1} \right)\left( {5 - 1 + 1} \right)}}{2} $ $ Total{\text{ }}no{\text{ }}of{\text{ }}the{\text{ }}spectral{\text{ }}lines\; = 10 $ So, the ten lines are; $ 5 \to 4,5 \to 3,5 \to 2,5 \to 1,4 \to 3,4 \to 2,4 \to 1,3 \to 2,3 \to 1,2 \to 1\; $ are possible in this case.Balmer series is displayed when electron transition takes place from higher energy states ( $ {{\mathbf{n}}_{\mathbf{h}}} = {\mathbf{3}},{\mathbf{4}},{\mathbf{5}},{\mathbf{6}},{\mathbf{7}}, \ldots $ ) to $ {{\mathbf{n}}_{\mathbf{l}}} = {\mathbf{2}} $ energy state. For Balmer series, $ {n_l} = 2 $ and $ {n_h} = 3,4,5 $ Thus, $ 5 \to 2,4 \to 2,3 \to 2\; $ it shows three lines lie in the visible region. Hence the final answer is three spectral lines lie in the visible region for the hydrogen atom when electrons jump from $ n_2 = 5\; $ to $ n_1 = 1\; $ in the visible region.

Note :

All the wavelength of the Balmer series falls in the visible part of the electromagnetic spectrum( $ 400nm{\text{ }}to{\text{ }}740nm $ ). Other series are lyman series, paschen series, bracket series, pfund series, Humphreys series. The series is observed at a higher wavelength. The spectral lines are extremely faint and widely spread out. They correspond to highly rare atomic events.

Maximum number of spectral lines emitted when electrons jump from n=5 to n=1 in hydrogen atom sample is 1) 3s 2) 2p 3) 2s 4) 1s

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How many spectral line will be observed when an electron jumps from n=5 to n=1 in the visible line spectra of H spectrum?

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