The figure below represents part of the emission spectrum for a one-electron ion in the gas phas

The figure below represents part of the emission spectrum...

Key Concepts

Emission Spectrum

An emission spectrum is the set of wavelengths of light emitted by an excited atom or molecule as electrons transition between quantized energy levels. Each line in the spectrum corresponds to a specific electronic transition, making it a powerful tool for identifying elements and studying atomic structure.

Electronic Transitions

Electronic transitions occur when an electron moves between discrete energy levels in an atom or ion. The energy difference between the levels is emitted as a photon, and the wavelength of the photon is determined by this energy gap, which is a fundamental concept in understanding atomic spectra.

Quantized Energy Levels

Quantized energy levels refer to the discrete values of energy that electrons can possess within an atom or ion. This quantization is a result of the wave nature of electrons and explains why atoms emit or absorb light at specific wavelengths, rather than a continuous spectrum.

Hydrogen-like Ions

Hydrogen-like ions are atomic systems that contain only one electron orbiting the nucleus, similar to the hydrogen atom. Their simplicity allows for accurate application of theoretical models, such as the Rydberg formula, to predict wavelengths of emitted or absorbed light during electronic transitions.

Rydberg Formula

The Rydberg formula is an equation used to predict the wavelengths of light resulting from electronic transitions in hydrogen-like atoms and ions. It relates the wavelength of emitted or absorbed light to the principal quantum numbers of the initial and final energy levels, utilizing fundamental physical constants.

Transcript

00:01 In this problem, we start off by being given some spectral lines, which i've abbreviated in this diagram here.

00:09 Each spectral line represents a transition from a different energy level to the final energy level of n equals three.

00:16 So as the wavelength increases, that means that our energy decreases or increases this direction, becomes more as you move away, which also means that the distance increases from the nucleus.

00:39 So if our final resting state is the n equals 3, the furthest line to the right would be at n equals 4.

00:49 So that makes b5 and a6.

00:53 So my spectral line for a corresponds to n equals 6, and my spectral line for b corresponds to n equals 6, equals five.

01:07 Once we have that information we can use it to identify the wavelengths of the different spectral lines.

01:16 And we do this by using the boers model for one electron atoms.

01:22 This tells us that the energy for any given transition is equal to a constant negative 2 .178 times 10 to the minus 18 times the ratio of z squared over nf squared minus z squared over n i squared z squared z is the atomic number and f is the final energy state which in this case is three and n i is the initial energy level which in this part of the problem we know is five because we're talking about line v we're also called the wavelength of this line.

02:39 And so we can use that information to find the energy because energy is equal to planck's constant times the speed of light divided by wavelength.

02:51 So we know that plank's constant is 6 .626 times 10 to the minus 34 times the speed of light or 3 times 10 to the 8, divided by the wavelength given, which is 142 .5 nanometers, or 142 .5 times 10 to the minus 9 meters.

03:20 Or the energy of this transition is equal to 1 .39 times 10 to the minus 18 tools.

03:38 And since this energy is being given off, it's going to be a negative value here.

03:43 So we can use that information to solve for our atomic number.

03:47 Of this atom.

03:48 So negative 1301 .39 times 10 to the minus 18 equals negative 2 .178 times 10 to the minus 18 times z squared which we don't know divided by 3 squared minus z squared over 5 squared.

04:17 I divided both sides by this number.

04:21 We'll have 0 .64 is equal to z squared over 9 minus z squared over 25.

04:53 And there are a bunch of different ways that you can solve this algebraically...

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