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General Physics (PHY 2140) Lecture 34 Modern Physics Atomic Physics De Broglie wavelength in the atom Quantum mechanics http://www.physics.wayne.edu/~apetrov/PHY2140/ Chapter 28 7/17/2015 1 Lightning Review Last lecture: 1. Atomic physics Bohr’s model of atom Ei E f hf mevr n , n 1, 2,3,... 1 1 RH 2 2 n f ni 1 Review Problem: Suppose that the electron in the hydrogen atom obeyed classical rather then quantum mechanics. Why should such an atom emit a continuous rather then discrete spectrum? 7/17/2015 2 Recall Bohr’s Assumptions Only certain electron orbits are stable. Radiation is emitted by the atom when the electron “jumps” from a more energetic initial state to a lower state Ei E f hf The size of the allowed electron orbits is determined by a condition imposed on the electron’s orbital angular momentum mevr n , n 1, 2,3,... Why is that? 7/17/2015 3 Modifications of the Bohr Theory – Elliptical Orbits Sommerfeld extended the results to include elliptical orbits Retained the principle quantum number, n Added the orbital quantum number, ℓ ℓ ranges from 0 to n-1 in integer steps 7/17/2015 All states with the same principle quantum number are said to form a shell The states with given values of n and ℓ are said to form a subshell 4 Modifications of the Bohr Theory – Zeeman Effect and fine structure Another modification was needed to account for the Zeeman effect The Zeeman effect is the splitting of spectral lines in a strong magnetic field This indicates that the energy of an electron is slightly modified when the atom is immersed in a magnetic field A new quantum number, m ℓ, called the orbital magnetic quantum number, had to be introduced m ℓ can vary from - ℓ to + ℓ in integer steps High resolution spectrometers show that spectral lines are, in fact, two very closely spaced lines, even in the absence of a magnetic field This splitting is called fine structure Another quantum number, ms, called the spin magnetic quantum number, was introduced to explain the fine structure 7/17/2015 5 28.5 de Broglie Waves One of Bohr’s postulates was the angular momentum of the electron is quantized, but there was no explanation why the restriction occurred de Broglie assumed that the electron orbit would be stable only if it contained an integral number of electron wavelengths 7/17/2015 6 de Broglie Waves in the Hydrogen Atom In this example, three complete wavelengths are contained in the circumference of the orbit In general, the circumference must equal some integer number of wavelengths 2 r n , 1, 2,3,... h , so me v but mevr n , n 1, 2,3,... 7/17/2015 This was the first convincing argument that the wave nature of matter was at the heart of the behavior of atomic systems 7 QUICK QUIZ 1 In an analysis relating Bohr's theory to the de Broglie wavelength of electrons, when an electron moves from the n = 1 level to the n = 3 level, the circumference of its orbit becomes 9 times greater. This occurs because (a) there are 3 times as many wavelengths in the new orbit, (b) there are 3 times as many wavelengths and each wavelength is 3 times as long, (c) the wavelength of the electron becomes 9 times as long, or (d) the electron is moving 9 times as fast. (b). The circumference of the orbit is n times the de Broglie wavelength (2πr = nλ), so there are three times as many wavelengths in the n = 3 level as in the n = 1 level. 7/17/2015 8 28.6 Quantum Mechanics and the Hydrogen Atom One of the first great achievements of quantum mechanics was the solution of the wave equation for the hydrogen atom The significance of quantum mechanics is that the quantum numbers and the restrictions placed on their values arise directly from the mathematics and not from any assumptions made to make the theory agree with experiments 7/17/2015 9 Problem: wavelength of the electron Determine the wavelength of an electron in the third excited orbit of the hydrogen atom, with n = 4. 7/17/2015 10 Determine the wavelength of an electron in the third excited orbit of the hydrogen atom, with n = 4. Given: Recall that de Broglie’s wavelength of electron depends on its momentum, = h/(mev). Let us find it, me vrn n , n=4 Recall that so mev n rn rn n2 a0 , so mev h 2 a0 n Find: e = ? 7/17/2015 Thus, h 2 a0 n 8 0.0529nm 1.33nm me v 11 Quantum Number Summary The values of n can increase from 1 in integer steps The values of ℓ can range from 0 to n-1 in integer steps The values of m ℓ can range from -ℓ to ℓ in integer steps 7/17/2015 12 QUICK QUIZ 2 How many possible orbital states are there for (a) the n = 3 level of hydrogen? (b) the n = 4 level? The quantum numbers associated with orbital states are n, , and m . For a specified value of n, the allowed values of range from 0 to n – 1. For each value of , there are (2 + 1) possible values of m. (a) If n = 3, then = 0, 1, or 2. The number of possible orbital states is then [2(0) + 1] + [2(1) + 1] + [2(2) + 1] = 1 + 3 + 5 = 9. (b) If n = 4, one additional value of is allowed ( = 3) so the number of possible orbital states is now 9 + [2(3) + 1] = 9 + 7 = 16 7/17/2015 13 Spin Magnetic Quantum Number It is convenient to think of the electron as spinning on its axis The electron is not physically spinning There are two directions for the spin Spin up, ms = ½ Spin down, ms = -½ There is a slight energy difference between the two spins and this accounts for the Zeeman effect 7/17/2015 14 Electron Clouds The graph shows the solution to the wave equation for hydrogen in the ground state The curve peaks at the Bohr radius The electron is not confined to a particular orbital distance from the nucleus The probability of finding the electron at the Bohr radius is a maximum 7/17/2015 15 Electron Clouds The wave function for hydrogen in the ground state is symmetric The electron can be found in a spherical region surrounding the nucleus The result is interpreted by viewing the electron as a cloud surrounding the nucleus The densest regions of the cloud represent the highest probability for finding the electron 7/17/2015 16 7/17/2015 17