(a) What is the smallest amount of energy in electron volts...

(a) What is the smallest amount of energy in electron volts that must be given to a hydrogen atom initially in its ground level so that it can emit the $H_{\alpha}$ line in the Balmer series? (b) How many different possibilities of spectral-line emissions are there for this atom when the electron starts in the $n=3$ level and eventually ends up in the ground level? Calculate the wavelength of the emitted photon in each case.

Key Concepts

Rydberg Formula

The Rydberg formula, 1/? = R (1/n_f² - 1/n_i²), relates the wavelengths of the emitted or absorbed photons to the initial (n_i) and final (n_f) quantum numbers. This formula is crucial for calculating the wavelengths of light emitted during electron transitions in hydrogen.

Cascade Transitions

When an electron in a higher excited state decays to the ground state, it can do so via multiple paths, emitting different photons along the way. Counting the possible transitions involves combinatorial reasoning, and each unique step (such as 3?2, 3?1, and 2?1) corresponds to a distinct spectral line.

Quantum Energy Levels

Hydrogen atoms have discrete energy states defined by quantum mechanics, with energy levels given by E? = -13.6 eV/n². Transitions between these levels, either by absorption of energy or by emission of photons, underlie much of atomic spectroscopy.

Balmer Series

The Balmer series consists of spectral lines that result from electron transitions where the final state is n = 2. The H? line is produced by the transition from n = 3 to n = 2, which is the most prominent line in this series and lies in the visible region.

Transcript

00:01 In this question, we are looking at the hydrogen atom.

00:04 We want to find what is the photon needed to excite a hydrogen atom at the ground state, such that you will emit the alpha line in the balmer series.

00:18 So the balmer series basically we are looking at transitions involving any excited state above 2 all the way down to 2.

00:26 So say for example 3 to 2, 4 to 2, 5 to 2, so on and so forth, right, as long as the ending state is in the end equals to 2 state.

00:38 So this is the balmer series.

00:42 So when we're looking at the alpha line of the balmer series, we are basically looking at the first possible transition with the smallest energy.

00:54 This is the case when we have the transition from 3 to 2 n equals 3 to n equals 2 that means that our hydrogen atom which is starting at the ground states must be first excited up to at least the second excited state which is the n equals 3 state before it is allowed to undergo this 3 to 2 transition therefore the the smallest amount of energy required to give to the hydrogen atom will be e3 minus e1.

01:36 Now we know that the energy levels of the hydrogen atom can be found by this general equation 13 .6ev divided by n square.

01:47 So this will be just 13 .6ev, 1 minus 1 over 3 square.

01:59 And this will give us 12 .1 ev.

02:04 So this is the minimum energy and minimum energy that must be given to this hydrogen atom to excite it to n equals to 3.

02:17 Now the next part we want to find how many different possibilities of emissions there are when you have this atom at the n equals to 3 state.

02:26 So now it's in the n equals 3 one of the possible photons that can emit before reaching to the ground level so one possible would be from 3 to 2, another possible would be from 3 all the way down to 1, and of course there's an intermediate step from 2 to 1...

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