Energy levels and spectra

In a nutshell

If electrons in energy levels are given energy, they can move to a higher energy level and become 'excited'. When they de-excite, they emit a photon of a specific wavelength, which can create a line on an emission spectrum.

Equations

Constants

Variables

Energy levels of electrons

Electrons bound to an atom only exist in discrete energy levels. Each element has its own set of energy levels. Each energy level is also negative as energy is required to remove the electron from the atom. Electrons absorb specific photons of energy to move up energy levels. This energy must be exactly the same as the difference between the two energy levels. When this happens electrons are said to be in an 'excited' state. 

When the electrons drop down energy levels towards the ground state, they emit a photon of energy with a specific frequency and wavelength, and are said to 'de-excite'. The ground state $$( E_0)$$​ is the lowest energy level of the atom.

Note: The energy levels are discrete as they only take on specific values. 

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Energy level diagrams

In the diagram below is the hydrogen balmer series. The energy levels are shown as numbers down the left hand side of the diagram, higher numbers representing higher energy. Each arrow represents a transition, the emission of a photon when an electron de-excites, as explained previously. The longer the line, the higher the energy of the transition,

1. Rising energy levels 2. Balmer series

Types of spectra

Line spectra are the wavelengths of light emitted from an element when the electrons lose energy, this energy is lost in the form of photons of a specific energy. 

Absorption line spectra

An absorption spectra is created when a continuous spectra is passed through a cool gas. The cool gas which could be made up of lots of different elements, will have the electrons in the ground state. 

When the continuous spectra is passed through the cool gas, the electrons will absorb photons of specific wavelengths and get excited. Only photons with the exact energy equal to the difference between the energy levels are absorbed, so only specific wavelengths are absorbed. 

These wavelengths create dark lines. When the white light is observed the other side of the gas, the photons which are absorbed are missing from the continuous spectrum.

Emission spectra

An emission spectra is created when an excited gas is allowed to cool. As the gas cools the electrons will start to de-excite and emit photons with specific wavelengths.

The diagram below shows what a continuous, emission and absorption spectra looks like.

1. Continuous spectrum 2. Emission spectrum 3. Absorption spectrum 4. Increasing wavelength

The emission lines correspond to the photons emitted and their wavelengths. The energy of the photon is described as:

​$$E= hf\newline \\[0.1in] E= \dfrac{hc}{\lambda}$$

where $$f$$ is the frequency of the photon, $$h$$ is the Planck's constant, $$c$$ is the speed of light and $$\lambda$$ is the wavelength.​

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Each energy level has a specific wavelength, and therefore you can work out which elements are in a gas as each element has a specific pattern which can be compared to a database on Earth. Below is an example of emission spectrums for hydrogen and helium.  

Example

An absorption line is observed in the spectrum of a particular gas when electrons absorb photons and move between energy levels at $$−9.1 \ eV$$​ and $$−3.2 \ eV$$ Calculate the wavelength the emitted photons.

Firstly, write down the known values:

​$$\Delta E = -9.1 -- 3.2 = -5.9 \ eV$$

Convert energy into $$J$$:

$$5 \ eV = 5 \times 1.6 \times 10^{-19} = 8 \times 10^{-19} \ J$$​​

Next write down the equation and rearrange for $$\lambda$$:

$$E = \dfrac{hc}{\lambda}\\[0.1in] \lambda = \dfrac{hc}{E}$$

Substitute values into the rearranged equation and solve: 

$$\lambda = \dfrac{6.63 \times 10^{-34} \times 3 \times 10^8}{8 \times 10^{-19}}\\[0.1in] \lambda = 2.486 \times 10^{-7} \ m \\[0.1in] \lambda = 2.5 \times 10^{-7} \ m \ \ (2sf)$$

The wavelength of the photons is $$\underline{2.5 \times 10^{-7} \ m}$$.​

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Diffraction gratings

Diffraction gratings have regularly spaced slits that diffract light. Different colours of light have different wavelengths, and will therefore be diffracted different amounts at different angles. The following equation can be used to determine the wavelength of light used:

​$$dsin\theta = n\lambda$$​​

where $$d$$ is the diffraction slit separation, $$\theta$$ is the angle of diffraction and $$n$$ is the order of maxima.

Example

Monochromatic light from a laser of wavelength $$634 \,nm$$​ is incident normally at a diffraction grating. The angle between the second-order maximum and the zero-order maximum is measured to be $$23°$$​. Calculate the grating spacing $$d$$.

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Firstly, write down the known values:

$$\lambda = 634\, nm\newline \theta =23 \degree\newline n=2$$

Next, write down the equation needed and rearrange for $$d$$​:

$$n\lambda = dsin\theta$$

$$d = \dfrac{n \lambda}{dsin\theta}\newline$$​​

Substitute values into the rearranged equation and solve:

$$d = \dfrac{2 \times 634 \times 10^{-9}}{sin(23)}\newline \\[0.1in]d= 3.3 \times 10^{-6}\,m$$

The grating spacing is $$\underline{3.3 \times 10^{-6}\,m}$$.

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