The maximum number of lines observed in a spectrum when a h-gas sample is excited to 4th state is

Uh-Oh! That’s all you get for now.

We would love to personalise your learning journey. Sign Up to explore more.

Sign Up or Login

Skip for now

Uh-Oh! That’s all you get for now.

We would love to personalise your learning journey. Sign Up to explore more.

Sign Up or Login

Skip for now

What is the maximum number of emission lines when the excited electron of an H atom in n = 6 drops to the ground state?

When the excited electron of an H atom in n = 6 drops to the ground state, the following transitions are possible:

Hence, a total number of (5 + 4 + 3 + 2 + 1) 15 lines will be obtained in the emission spectrum.

The number of spectral lines produced when an electron in the nth level drops down to the ground state is given by `("n"("n"-1))/2`

Given, n = 6

Number of spectral lines =` (6(6-1))/2 = 15`

Concept: Bohr’s Model for Hydrogen Atom

  Is there an error in this question or solution?

Answer

Hint: There is a formula to calculate the number of spectral lines or emission lines formed by the electrons when it drops from orbit ‘n’ to ground state and it is as follows.\[\text{Number of spectral lines = }\dfrac{n(n-1)}{2}\] Where n = orbit from where the electron starts to drop to ground state.

Complete step by step answer:

- Means an electron forms 15 emission lines when it drops from n = 6 to ground level.

Note: When the electron drops from higher orbit to lower orbit means the electron losing or emitting energy then the electron forms some lines related to emitted energy then they are called as emission lines or spectral lines. When an electron jumps to a higher energy level then it absorbs some energy and forms an absorption spectrum or absorption spectral lines.

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 $N$.