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# Hydrogen Spectra explained

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• 1. Explanation of spectra In 1885 Johann Balmer, a Swiss physicist, discovered, by trial and error, that the energies in the emission spectrum of hydrogen were given by the formula: Johann Balmer ⎛1 1⎞ (1825-1898) ∆E = −Rz gZ ⎜ 2 − 2 ⎟ 2 ⎝ n f ni ⎠ where n may take integer values 3, 4, 5, … and R is a constant number
• 2. Explanation of spectra Since the emitted light from a gas carries energy, it is reasonable to assume that the emitted energy is equal to the difference between the total energy of the atom before and after the emission. Since the emitted light consists of photons of a specific wavelength, it follows that the emitted energy is also of a specific amount since the energy of a photon is given by: hc E = hf = λ This means that the energy of the atom is discrete, that is, not continuous.
• 3. The “electron in a box” model If the energy of the atom were continuous the emission of light wouldnt always be a set of specific amounts. The first attempt to explain these observations came with the “electron in a box” model. Imagine that an electron is confined in a box of linear size L. If the electron is treated as a wave, it will have a wavelength given by: h λ= the electron can only be found somewhere p along this line x=0 x=L
• 4. The “electron in a box” model If the electron behaves as a wave, then:  The wave is zero at the edges of the box  The wave is a standing wave as the electron does not lose energy This means that the wave will have nodes at x=0 and x=L. This implies that the wavelength must be related to the size of the box through: 2L λ= n Where n is an integer
• 5. The “electron in a box” model Therefore, the momentum of the electron is: h h mh p= = = λ 2L 2L n The kinetic energy is then: 2  mh  2   2 2 p  2L  = n h Ek = = 2m 2m 8mL2
• 6. The “electron in a box” model This result shows that, because the electron was treated as a standing wave in a “box”, it was deduced that the electron’s energy is quantized or discrete: h2 1× n =1 8mL2 h2 Ek = 4× 8mL2 n=2 h2 9× n=3 8mL2 However, this model is not correct but because it shows that energy can be discrete it points the way to the correct answer.
• 7. The Schrödinger theory In 1926, the Austrian physicist Erwin Schrödinger provided a realistic quantum model for the behaviour of electrons in atoms. The Schrödinger theory assumes that there is a wave associated to the electron (just like de Bröglie had assumed) Erwin Schrödinger This wave is called wavefunction and (1887-1961) represented by: ψ( x, t ) This wave is a function of position x and time t. Through differentiation, it can be solved to find the Schrödinger function: ∂ 2i ψ( r , t ) = − ∇ ψ( r , t ) +V ( r ) ψ( r , t ) 2 ∂t 2m
• 8. The Schrödinger theory The German physicist Max Born interpreted Schrödingers equation and suggested that: 2 ψ ( x, t ) can be used to find the probability of finding an electron near position x at time t. This means that the equation cannot tell exactly where to find the electron. This notion represented a radical change from classical physics, where objects had well-defined positions.
• 9. The Schrödinger theory Solving for Hydrogen, it is found that: 13.6 E = − 2 eV n In other words, this theory predicts that the electron in the hydrogen atom has quantized energy. The model also predicts that if the electron is at a high energy level, it can make a transition to a lower level. In that process it emits a photon of energy equal to the difference in energy between the levels of the transition.
• 10. The Schrödinger theoryBecause the energy of the photon is given by E = hf, knowing theenergy level difference, we can calculate the frequency andwavelength of the emitted photon.Furthermore, the theory also predicts the probability that aparticular transition will occur.This high n is essential to energy energyunderstand why levels very 0 eV close tosome spectral lines each otherare brighter than n=5others. n=4 n=3Thus, theSchrödinger theory n=2explains atomic -13.6 eVspectra. n=1