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Atomic physics PHY232 Remco Zegers

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1 Atomic physics PHY232 Remco Zegers
Room W109 – cyclotron building

2 lon-capa 27,28,29 last homework set: chapter 27: problems 1-10
PHY232 - Remco Zegers -

3 quiz (extra credit) If the photoelectric effect is observed in one metal, can you conclude that the effect will also be observed in another metal under otherwise the same conditions? a) Yes b) No answer: no. The workfunctions are different PHY232 - Remco Zegers -

4 models of the atom Newtonian era: the atom is a solid sphere
Thompson (~1900): atom is a sphere of positive charge with negatively charged electrons in it Rutherford (1911) devised a ‘planetary model’ with a positively charged nucleus in the core and electrons orbiting around it. where do atomic transitions come from continuous centripetal acceleration and thus emission of photons Bohr (1913): model for Hydrogen(-like) atoms (we’ll look at it) Quantum-mechanical description of the atom PHY232 - Remco Zegers -

5 PHY232 - Remco Zegers -

6 atomic spectra if an Hydrogen atom gets ‘excited’ (for example by heating it), light is emitted of certain specific wavelengths following the equation: 1/=RH(1/22-1/n2) RH=1.097x107 m-1 n=3,4,5,6... Light of these wavelengths gets absorbed if white light (consisting of all wave lengths) is shone on a Hydrogen gas. Balmer series, after its discoverer Johann Balmer PHY232 - Remco Zegers -

7 atomic spectra Similar spectra are observed for other elements, but the patterns are more complicated. Nevertheless, measuring such spectra allows one to identify which elements are present in a sample PHY232 - Remco Zegers -

8 Bohr’s model for Hydrogen
proton electron PHY232 - Remco Zegers -

9 Bohr’s theory for Hydrogen
assumption 1: the electron moves in circular orbits around the proton. The Coulomb force between the + nucleus and the – electron produces the centripetal acceleration. As a result, one can deduce the kinetic energy of the electron proton 1 Note: this is a pure classical reasoning PHY232 - Remco Zegers -

10 Bohr’s theory for Hydrogen
assumption 2: only certain orbits are stable, namely those in which no electromagnetic radiation is emitted in the absence of external forces. Hence, the energy of the atom is constant and the equations on the previous slide can be used. The size of the allowed orbits are those for which the orbital angular momentum of the electron is a multiple times ħ (h/(2)): This is derived (de Broglie) from the assumption that a fixed number of electron-wavelengths must fit in the orbit: 2 n=1 n=2 n=3 PHY232 - Remco Zegers -

11 n=7 PHY232 - Remco Zegers -

12 Bohr’s theory for Hydrogen
assumption 3 Radiation is emitted when an electron jumps from an outer orbit to an inner orbit. The energy of the radiation (and thus the frequency) is determined by the change in the atom’s energy due to the jump radiation is absorbed if an electron jumps from an inner orbit to an outer orbit The energy of an orbit can be calculated with: E=Ekinetic+Epotential 4 3 PHY232 - Remco Zegers -

13 Bohr’s theory for Hydrogen
combining assumptions 1,2 & 3: 1 & 2 give (solving for r, while eliminating v) combine with 3 3 1 2 5 PHY232 - Remco Zegers -

14 Bohr’s theory for Hydrogen
finally, by combining 4 and 5 and using c=f and RH=1.097x107 m-1 4 5 6 with nf and ni : integers >0 PHY232 - Remco Zegers -

15 The hydrogen spectrum emission spectrum absorption spectrum
By measuring the wave length of the light, one can determine the energy spectrum of Hydrogen n=1: ground state (energy is –13.6 eV) n=: electron is removed from atom: the atom is ionized. The ‘n’ is usually referred to as a shell; the 1st shell, the 2nd shell etc PHY232 - Remco Zegers -

16 question The quantum number n can increase without limit.
Therefore, the frequency of the emitted light from state n to the ground state can also increase without limit. true false answer: b) if nf grows, f grows asymptotically to cRH or if the frequency would grow without limit, the energy involved would also increase without limit, which doesn’t make much sense PHY232 - Remco Zegers -

17 example 1 How much energy does it take to ionize a Hydrogen atom?
answer: to ionize the atom, light must be absorbed. The initial state has n=1 (ground state) The final state has n= E1=-13.6 eV E=0 eV therefore, 13.6 eV must be absorbed by the Hydrogen atom PHY232 - Remco Zegers -

18 example 2 What is the wavelength of the light emitted if an electron goes from the 5th shell to the 2nd shell in a Hydrogen atom? What is the energy of the photon? answer: in the initial state: n=5 E5=-13.6/52= eV in the final state: n=2 E2=-13.6/22=-3.4 eV Ei-Ef= =2.846 eV 1/=RH(1/22-1/52)=1.097x107(1/4-1/25)=2.30x106 m-1 =434 nm PHY232 - Remco Zegers -

19 example 3 What is the wavelength of the light absorbed if a hydrogen atom in its ground state is excited into its n=4 state? How much energy is absorbed (what is the excitation energy)? answer: in the initial state: n=1 E1=-13.6/12=-13.6 eV in the final state: n=4 E4=-13.6/42=-0.85 eV Ei-Ef= = eV (note: negative so absorbed) 1/=RH(1/12-1/42)=1.097x107(1/1-1/16)=1.03x107 m-1 =97.2 nm note: I switched Ef and Ei in the equation for 1/, since that equation was derived for excitation and not for absorption PHY232 - Remco Zegers -

20 Heavier atoms Bohr’s equation also does well for heavier atoms IF they have been ionized such that only one electron remains in its orbit. For example for Helium (2 protons (Z=2) in the nucleus), 2 electrons in the orbits would make it neutral, but only if one is missing can Bohr’s equations be applied. The equations need to be slightly modified however, to take into account that the Coulomb forces/energies are different. Change e2 into Ze2 everywhere where it occurs. Z2 5a Z2 6a PHY232 - Remco Zegers -

21 question. Consider a hydrogen atom and a singly ionized helium atom. Which one has the lower ground state energy? a) hydrogen b) singly ionized helium c) the same Z2 answer: b) in both cases n=1, but EHe+= -13.6x4 eV and EH+=-13.6 eV. One can also argue that the attractive Coulomb force is stronger! PHY232 - Remco Zegers -

22 More general description
Bohr’s classical approach breaks down if more than 1 electron is present in the atom. Instead, the problem has to be treated quantum mechanically by solving the Schrödinger equation The solutions give the distributions of electron clouds (so-called wave-functions) in the atom. The clouds describe the probability of finding an electron in a certain positions. The clouds are characterized by quantum numbers, which follow ‘simple’ rules the wave functions for Hydrogen PHY232 - Remco Zegers -

23 atomic shells and quantum numbers
The electrons are ordered according to 4 quantum numbers the principal quantum number n range: 1,2,3…. Usually referred to as K(n=1), L(n=2), M(n=3), N(n=4) shells the orbital quantum number l range: 0,1,2,…n-1 (so there are n possibilities) usually referred to as s (l=0),p (l=1) d (l=2) ,f (l=3),g (l=4),h,I… the orbital magnetic quantum number ml range: -l, -l+1,-1,0,1…l-1,l (there are 2l+1 possibilities) the spin magnetic quantum number ms range: -1/2 (electron spin up) or + ½ (electron spin down) in each state with given n,l,ml one can maximally place 2 electrons (ms=-1/2 and ms=+1/2). Pauli exclusion principle no two electrons in an atom can have the same set of quantum numbers n,l,ml,ms PHY232 - Remco Zegers -

24 example: Hydrogen Hydrogen (not ionized) has only 1 electron.
ground state quantum numbers: n=1 l=0,…,n-1=0,0 so only l=0 ml=-l,…,+l so only ml=0 ms=-1/2 or +1/2 one could maximally place 2 electrons in here (different ms) this level is referred to as 1s1: one electron in the level with quantum numbers n=1, l=0 (s) n=2 states n=2 l=0,n-1=0,1 so l=0 or l=1 ml=-l,…+l, so ml=0 if l=0 and ml=-1,0,1 if l=1 for each ml, ms=-1/2 ot +1/2 these levels are referred to as: 2s0 : n=2, l=0 it is empty but I could put 2 electrons in there 2p0: n=2, l=1 it is empty but I could put 6 electrons in there, namely: two each in n=2,l=1 with ml=-1,0,1 PHY232 - Remco Zegers -

25 more complicated example, Sodium Na:
Sodium has Z=11 (11 protons), so if not ionized, it has 11 electrons. Atomic level will fill up according to lowest n, then lowest l there are exceptions to this (see also table in back of book and table 28.4) n=1, l=0, ml=0, ms=-1/2,+1/2 1s2 2 electrons n=2, l=0, ml=0, ms=-1/2,+1/2 2s2 2 electrons n=2, l=1, ml=-1,0,1, ms=-1/2,+1/2 2p6 6 electrons n=3, l=0, ml=0, ms=-1/2,+1/2 3s1 1 electron (2 possible, but only 1 needed to get to 11) SUM: electrons So, the ground state configuration can be described as: 1s22s22p63s1 Note that the n=1 and n=2 shells are filled with 10 electrons (Neon) so this is sometimes referred to as: [Ne]3s1 PHY232 - Remco Zegers -

26 energy levels (example for Li Z=3)
electrons outside the last filled (sub)shell are most important for the chemical properties of an atoms. PHY232 - Remco Zegers -

27 questions a) what is the electron configuration of Argon (Z=18)
b) what is the electron configuration of Titanium (Z=22) start filling: 1s2 n=1, l=0,ml=0,ms=-1/2 +1/2 2 electrons 2 2s2 n=2, l=0, ml=0,ms=-1/2 +1/2 2 electrons 4 2p6 n=2, l=1, ml=-1,0,1,ms=-1/2 +1/2 6 electrons 10 3s2 n=3, l=0, ml=0,ms=-1/2 +1/2 2 electrons 12 3p6 n=3, l=1, ml=-1,0,1,ms=-1/2 +1/2 6 electrons 18 The 3p sub-shell is just filled b) start with Argon and add 4 electrons [Ar]+ 3d4 n=3, l=2,ml=-2,-1,0,1,2 ms=-1/2 +1/2 4 electrons 22 only 4 out of 10 are needed Note: in reality the 4s shell gets filled before the 3d shell, so the real ground-state configuration is [Ar]4s23d2 (see table 28.4) PHY232 - Remco Zegers -

28 For Titanium the 4s level has moved down and the 3d level moved up, so that the 4s level is lower in energy than the 3d level PHY232 - Remco Zegers -

29 Periodic table of elements
see also back of book Periodic table of elements Filled sub-shell, so very stable ‘Noble gases’ alkali metals very reactive one loosely bound electron halogens: very reactive, need one more electron to fill shell PHY232 - Remco Zegers -

30 periodic table structure
PHY232 - Remco Zegers -

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