Atomic physics PHY232 Remco Zegers [email protected] Room W109 – cyclotron building...

Atomic physics

PHY232Remco [email protected] W109 – cyclotron buildinghttp://www.nscl.msu.edu/~zegers/phy232.html

PHY232 - Remco Zegers - 2

lon-capa 27,28,29

last homework set:

chapter 27: problems 1-10chapter 28: problems 11-14chapter 29: problems 15-17


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


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


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


atomic spectra

Similar spectra are observed for other elements, but the patterns aremore complicated. Nevertheless, measuring such spectra allows oneto identify which elements are present in a sample


Bohr’s model for Hydrogen

proton

electron


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

Note: this is a pure classical reasoning

1

proton


Bohr’s theory for Hydrogenassumption 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:

n=1 n=2 n=3

2


n=7



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

3

4


combining assumptions 1,2 & 3:

1 & 2 give (solving for r, while eliminating v)

combine with 3


1 23

5


finally, by combining 4 and 5

and using c=f and RH=1.097x107 m-1


4 5

with nf and ni : integers >0

6


The hydrogen spectrum

emission spectrum

absorption spectrum

By measuring the wave lengthof the light, one can determinethe 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


question

The quantum number n can increase without limit. Therefore, the frequency of the emitted light from staten to the ground state can also increase without limit.a) trueb) 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’tmake much sense


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 eVE=0 eVtherefore, 13.6 eV must be absorbed by the Hydrogen atom


example 2What 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=-0.544 eVin the final state: n=2 E2=-13.6/22=-3.4 eVEi-Ef=-0.554+3.4=2.846 eV1/=RH(1/22-1/52)=1.097x107(1/4-1/25)=2.30x106 m-1

=434 nm


example 3What 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 eVin the final state: n=4 E4=-13.6/42=-0.85 eVEi-Ef=-13.6+0.85=-12.75 eV (note: negative so absorbed)1/=RH(1/12-1/42)=1.097x107(1/1-1/16)=1.03x107 m-1

=97.2 nmnote: I switched Ef and Ei in the equation for 1/, since that equationwas derived for excitation and not for absorption


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 Z2

5a

6aZ2


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 Z2

answer: b) in both cases n=1, but EHe+= -13.6x4 eVand EH+=-13.6 eV.One can also argue that the attractive Coulomb forceis stronger!


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


atomic shells and quantum numbersThe 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


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


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: 11 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


energy levels (example for Li Z=3)

electrons outside the last filled (sub)shell are most importantfor the chemical properties of an atoms.


questions

a) what is the electron configuration of Argon (Z=18) b) what is the electron configuration of Titanium (Z=22)

a) start filling:1s2 n=1, l=0,ml=0,ms=-1/2 +1/2 2 electrons 22s2 n=2, l=0, ml=0,ms=-1/2 +1/2 2 electrons 42p6 n=2, l=1, ml=-1,0,1,ms=-1/2 +1/2 6 electrons 103s2n=3, l=0, ml=0,ms=-1/2 +1/2 2 electrons 123p6 n=3, l=1, ml=-1,0,1,ms=-1/2 +1/2 6 electrons 18The 3p sub-shell is just filledb) 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 neededNote: in reality the 4s shell gets filled before the 3d shell, so the real ground-state configuration is [Ar]4s23d2 (see table 28.4)


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


Periodic table of elementsFilled sub-shell,so very stable‘Noble gases’

alkali metalsvery reactiveone loosely boundelectron

halogens: very reactive, need one more electron to fill shell

see also back of book


periodic table structure

Atomic physics PHY232 Remco Zegers [email protected] Room W109 – cyclotron building...

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