Download n=1 l=0

Atomic physics
PHY232
Remco Zegers
[email protected]
Room W109 – cyclotron building
http://www.nscl.msu.edu/~zegers/phy232.html
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
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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
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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
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Bohr’s model for Hydrogen
proton
electron
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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
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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)):
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
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n=7
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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
4
 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 Ekin:
3
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Bohr’s theory for Hydrogen
combining assumptions 1,2 & 3:
1
2
3
1 & 2 give (solving for r, while eliminating v)
combibe with 3
5
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Bohr’s theory for Hydrogen
finally, by combining 4 and 5
4
5
and using c=f and RH=1.097x107 m-1
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with nf and ni : integers >0
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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
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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.
a) true
b) false
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example 1
How much energy does it take to ionize a Hydrogen atom?
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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?
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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)?
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lon-capa
do problem 2 from lon-capa 12
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Heavier atoms
Bohr’s equation also does well for heavier atoms IF they have been
ionized such that only one electron remains in its orbits.
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
6a
Z2
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lon-capa
 do problem 1 from lon-capa 12
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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
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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
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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
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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
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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
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energy levels (example for Li Z=3)
electrons outside the last filled (sub)shell are most important
for the chemical properties of an atoms.
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questions
 a) what is the electron configuration of Argon (Z=18)
 b) what is the electron configuration of Titanium (Z=22)
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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
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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
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periodic table structure
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lon-capa
 do problem 3 from lon-capa 12.
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