Lecture 4. The Bohr model of the atom. De Broglie theory. The Davisson-Germer experiment
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1 Lecture 4 The Bohr model of the atom De Broglie theory The Davisson-Germer experiment
2 Objectives Learn about electron energy levels in atoms and how Bohr's model can be used to determine the energy levels in hydrogen. Learn about the inadequacies of the Bohr model. Introduce de Broglie's theory that all matter has a characteristic wavelength.
3 Rutherford backscattering experiment Helium nuclei are fired at a gold foil. Some of the nuclei are backscattered they must be colliding with a dense, heavy object within the gold atoms. Rutherford realised that the atomic nucleus contained protons and neutrons, with the electrons orbiting the nucleus. The nucleus is only about 1/ th the size of the atom, so most of the atom consists of empty space.
4 Classical image of an atom The electrons are pictured as a particle in orbit about the nucleus of an atom. If this were really the case, centripetal forces would soon pull the electrons into the nucleus and we would see a release of energy in a continuous spectrum.
5 Emission spectra In fact, when an atom is excited we find energy is released at discrete frequencies and the electrons never end up in the atomic nucleus. Put hydrogen gas in a tube and pass a current through it. The hydrogen gives off light with a characteristic spectrum. Visible spectral lines for hydrogen Bohr suggested that electrons live in stable orbits, each with a defined energy level. When the electrons undergo transitions between levels the energy absorbed or emitted is quantised into discrete wavelengths. The electrons do not emit radiation when they are in a stable orbit.
6 Balmer s formula n2 Balmer came up with the formula = to account for the hydrogen n 2 4 spectrum. It was in excellent agreement with experimental results. If we put n = 3, 4, 5 we get the wavelengths of the visible spectrum of Hydrogen, the so-called Balmer series.
7 Bohr s model Bohr developed his model of the hydrogen atom in Postulate 1 : Hydrogen exists in discrete energy states. These states are characterised by discrete values of angular momentum. In these states the atom does not radiate energy : L=nħ=mvr, L is the angular momentum. Postulate 2 : When an atom undergoes a change in energy from E n to E m, electromagnetic radiation is given off at a frequency of hf=e n -E m. We start by setting the Coulomb force equal to the centripetal force, i.e. e r = mv2 2 r of free space.,where e is the charge on an electron and ε 0 the permittivity Now, v=nħ/mr and Z=1/4πε 0 ( m/f) so Ze2 r = m n2 ħ 2 and r= n2 ħ 2 2 m 2 r 3 me 2 Z r=a 0 n 2. Where a 0 =ħ 2 /me 2 Z, a 0 is known as the Bohr radius ( m). or
8 Bohr s model II The energy of the n th orbit is given by the sum of the kinetic and potential energies, i.e. E=K +U= 1 2 mv 2 Ze2 r Substitute for v : E= 1 2 m n ħ 2 Ze 2 mr r So E= me4 Z 2 2 n 2 ħ 2 me4 Z 2 n 2 ħ = me 4 Z n 2 ħ 2 and r : E= 1 2 m n ħ m e2 Z m n 2 ħ 2 2 Ze 2 me2 Z n 2 ħ 2 Let R=me 4 Z 2 /2ħ 2, the Rydberg constant = 13.6 ev. (Ionisation potential). For transitions between two orbits n 1 and n 2 we have E 12 =R 1 n n. 2
9 Finite mass correction The above presumes that the mass of the atomic nucleus is infinite. If we consider the real mass of the atomic nucleus we need to replace the value of m with m= m e M n m e M n, where M n is the mass of the atomic nucleus. m e = kg (Electron) m p = kg (Proton) m n = kg (Neutron) For hydrogen the nucleus consists of just one proton so m= kg, For deuterium the nucleus consists of a proton and a neutron so m= kg. This small difference translates into a measurable difference in the position of the spectral lines of deuterium. It was through this effect that deuterium was discovered.
10 De Broglie s theory In 1924 de Broglie suggested that all matter should display wavelike behaviour. The wavelength of any object is given by λ=h/p. For everyday objects λ is very small, but for electrons it isn t. If electrons behave as waves it is easier to imagine how their atomic orbits are restricted to multiples of h. They must form a kind of standing wave around the nucleus. Standing wave orbitals
11 Standing waves Suppose an electron, having momentum p, is moving in a circular orbit of radius r. Then for a standing wave, a whole number of wavelengths must fit around the circle, so for some integer n, n =2 r. If =h/ p we get 2 r=nh/ p or pr=nh/2 =n ħ. The angular momentum is L=mvr= pr=n ħ. By assuming a standing wave we obtain Bohr's first postulate of quantised angular momentum. For a free electron E= 1 2 mv 2 = p2 2 m, i.e. p= 2 m E and = h 2 m E.
12 Failures of the Bohr model The Bohr model is excellent at explaining the frequencies of the spectral lines obtained from hydrogen. But there are some effects which are not covered by the model. These include: The brightness of spectral lines the model does not address the problem of calculating the probability of a certain transition. In the model we are assuming a fixed radius for the electron orbits. According to Heisenberg s uncertainty principle this is not allowed we cannot know the precise radius of an orbit. The restriction to circular orbits is also unrealistic. The model cannot be extended to systems with more than one electron.
13 Davisson-Germer experiment This experiment showed that electrons have wavelike properties when scattered from a Ni crystal. de Broglies theory was confirmed.
14 Davisson-Germer experiment II Bragg s law n =2 d sin B. d=d sin / 2, =50 θ B =65 =90 /2 sin θ B =sin 90 /2 =cos /2 So n =2 D sin / 2 cos / 2 n =D sin where D is the atomic spacing From de Broglie =h/ p and p= 2 m E so D= nh 2 m E sin. For =50 and E=54 ev we get D = 2.179Å. (Actually D Ni = 2.15Å).
15 Conclusions The Bohr model of the atom gives an accurate description of the energy levels in hydrogen. For other atoms and electrons with non-spherical orbitals we need to use the Schrödinger equation to accurately calculate the energy levels. In solids and high pressure gases the wavefunctions of the electrons can overlap, leading to the creation of energy bands.
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