4b Flashcards
whats H made up of
1 proton
1 electron
in a hydrogen,, how do we define where the e- is in respect to the proton
we draw a +1 dot for the proton and draw a ‘x’ , ‘y’, and ‘z’ axis coming out of it.
the z axis is going up,, the y axis is coming towards us and the x axis is going to the rhs
the e- we then draw in its position
proton is at the origin
H’ = what
T’ + V’
potential and kinetic energy
H’ = what more specific: energy of the system!!: total way the system has energy
H’ = T’.nuc + T’.el + V’. nuc-el
bc the nuc can move and so has kinetic energy
the e- can move and have kinetic energy
the potential energy between the nuc and the e-
how the nuc + el move and energy between those 2 things.
in the H atom when we draw the 3 line axis wirth the nuc and the electron,,, can the potential energy,, V’ ,, equal to 0
nope
its no longer 0 bc its the nergy between the nuc and the electron
and bc they have opposite charges they have potential energy.
what does V’ nuc-el mean and how does it change
energy between nuc and electron bc theyre oppositely charged this isnt 0.
the distance between the nuc and the electron alters the V’. nuc-el.
bc the nuc is so heavy,, what does the e- think about it
the e- think its stood still
bc the e- moves so fast and the proton moves so slow the e- thinks its not moving.
e- just follows the proton / nuc.
bc the nuc barely moves bc its so much heavier (made up of 1 proton) than the e- ,, in hydrogen,, what can we say about H’
we said H’ = T’.nuc + T’.el + V’.nuc-el
but bc we just said the nuc is sm heavier and so it looks like its not moving,, we can say the
H’ = T’.el + V’.nuc-el
nuc T’ has a tiny tiny effect so we can ignore its Ke. we just need to think about where the e- is.
born oppenheimer approx
basically saying how H’ is approx equal to H’ = T’el + V’.nuc-el
due to nuc moving so slowly - we can ignore its T’
so using the born oppenheimer approx,, how can we write the schrodinger equation:
HY = EY for a H atom.
(V’nuc-el + T’. el)Y = EY
how many directions can the e- move in
the e- can move in 3 different directions
so we need a 3Dimensional
okay so if movement of an e- is now 3D,, what do we need to change
what T’.el is equal to
T’.el = - (h.dash^2/2me x V-^2)
where m = mass of an e-.
V-^2 = del squared = d2/dx2 + d2/dy2 + d2/dz2
= 3d version due to moving e- in more directions.
for the H atom,, is V’ = 0
nopeeee
why isnt V’=0 for the Hatom
bc theres potential between the nuc and the e- bc theyre opposite charges.
what is V’ for the hydrogen atom
= -Ze^2 / 4nEo x 1/r
Z = nuc charge
e = charge on e-
n = pi
Eo = vacuum permittivity (amount of screening//screaming done between the nuc and the e-)
r = electron - nuc distance.
the charge is a constant
dont rememeber this -
1/r = distance
Z for H =
1
Z for He =
2
what does the V’.nuc-el acc depend on
the variable part of the equation we dont need to memorise
aka 1/r
where r is the distance between the nuc and the e-.
this is the only thing that changes the V’.nuc-el.
distance getting larger = smaller V.
distance getting closer = larger V
isotropic meaning
same in all directions
V’ is the same in all directions : as long as the distance is the same
when do we use spherical // polar coordinates
when a value is isotropic aka it doesnt change with direction //angle.
instead of Y(x,y,z) we use Y(r,o|, 0-)
r = distance
other things = 2 angles.
which hamiltonian operator can we find // try to find now that we know V is isotropic
we try find the Y(r,o|, o-) one
we can find it bc we can split it up into radial part and angular part.
what can we separate the isotropic wavefunction into Y(r,o|,o-)
radial part: R(r)
angular part: Y(o|, o-)
this is not an approx!!
we can figure out each one separately.!! due to using the polar spherical coordinatesssss
okay so now that we can separate the angular and radial parts of the wavefunction,, how does this effect the schrodinger equation
we now get 2 schrodinger equation:
H’.rad Y.rad = E.rad Y.rad
radial schrodinger. energy we get from moving e- away from nuc!!! solved via analytical solutions: we can find a radial hamiltonian, find eiganfunctions and eigenvalues!!
and the angular one aswelllll
Y. rad looks like what
-e^ beta.r
or smt like that
looks like an exponential!!
Y.rad against r graph gives what
gives an exponential!
e^-Br (beta)
where r is the distance between the nuc and the e-.
r is always larger or equal to 0
so its an eigenfunction,, finite, normalised, single valued and continuous
its literally just a slope down
what form do radial wavefunctions have + what are solutions dependent on + how are these expressed
R(r) = (polynomial in r ) x e^-Br
solutions are dependent on the angular momentum ,, expressed in the Laguerre polynomial
have we solved the radial schrodinger equation
yes!!
we found that Y.rad = e^-Br
weve solved it and found eigenfunctions
we then get eigen values which give us energy!!
what gives us E in schrodinger equation
the eigenvalues
steps for schrodinger
find eigenfunctions
then find eigenvalues as these give us energy
energies for radial solutions =
Ec,,
- bunch of constants that give the rhydberg constant x 1/n^2
where n = integer, 1,2,3 (principal quantum number)
energy only depends on this!!
so energy = - rhydberg / n^2
okay so what have we done so far
we want to solve the full schrodinger equation: H’Y=EY
so we broke it down into Y(r, o|,o-)
= R(r) radial and Y(o|,o-) angular parts
weve solved an radial schrodinger equation: H’.rad Y.rad = E.rad Y.rad
to get radial wavefunction: e^-Br and energies (eigenvalue)
these help explain whats going on in a hydrogen atom
when u get sphercal polar coordinates,, what other force is occuring
centrifugal force bc its spinning around the proton// nuc aswell.
different values of ‘n’ will give what
different wavefunctions and different energies
diff principle number
n = 1 wavefunction radial isssss
e^-Xr
e^-something
starts high up and just slopes down
with Y as y axis and r as x axis. r being distance from nuc to e-
n =2 wavefunction shape radial
starts high,, dips below x axis,, is modified by the polynomial and the laguerr polynomial and then rises but still below x axis
n = 3 wavefunction radial
starts high,, dips below x axis
gets back ubove x axis but stays close to it
exponential that is modified by a different polynomial
all energies areeee
negative
bc Energy = - Rh/n^2
the most negative one is n=1 (ground state)
n=2 is higher(less negative) and n = 3 is even higher(less negative)
when looking at the radial wavefunction graphs for the different n values,, what should we look out for
the x axis unitssss
look at the range of different values!!
low energy way of arranging an e- around a proton =
n = 1
bc energy is based on n: ground state
also shows how the e- spends most of its time near the nucleus: bc Y is high at a lower value of ‘r’.
the larger the n value,, the principle quantum number,, the higher energy way of arrangement it is. (arranging an e- around a proton)
larger n valueeeee(radial) in terms of energy
the closer to 0 it is(less and less negative.,,, closer to positive)
angular schrodinger equation
H’.ang Y.ang = E.ang Y.ang
what is the angular schrodinger
spherical haronics
used as solutions for when quantum particles move around smt // around an origin
aka the e- around the proton: its not free to go anywhere,, its quantised according to spherical harmonics
what do we need to describe the angular motion of the e- around the central proton
angular momentum , l
and magnetic quantum number, ml
so u get Y.ang(l, ml)
u need 2 quantum numbers to determine the angular motion of the e-
spherical harmonics give rise to what
they give rise to orbital shapes , s, p ,d orbital shapes.
l = 0 =
s
l = 1
p
l = 2
d
what does l do ,, angular quantum number + what values can it have
shape of angular motion
0 –> n-1
ml = + what values can it have
orientation around the nucleus !!
same shape tho
bc the angular motion is quantised
- l —> l
why are there 3 different p orbitals
bc p : l = 1
so ml : -1, 0 1
so 3 diff ml’s so 3 different p orbitals
solving the angular schrodinger equation gives ussss
info about the different orbitald and their orientation: all based on one e- moving around a proton.
gives us spherical harmonics (solution to angular schrodinger equation)
we get different ways of rearranging an e- around the proton but they all have the same energy
spherical harmonics show us theeee
shape of the orbital
E.ang
constant
all have the same energy
what is an orbital
a one electron wavefunction
H’ Y(1e-) = E Y(1e-)
and that wavefunction is an orbital
and that equals the radial part x the angular part (remember that we split this up to get 2 different schrodinger equations and solved them separately)
Y.radial needs
‘n’
Y.angular needs
l , ml
so the wavefunction we need for 1e- isssss
Y(1e-)
product of the 2 split up ones: aka the angular x radial
so u have Y(n,l,ml)
we need these 3 to say which wavefunction
n: tells us how its moving radially,, relative to the nuc
ml, l : tells us how its moving angularly, relative to the nucleus
we need all 3 to tell us what an orbital looks like.
what do we need to tell us what an orbital looks like
Y(n,ml,l)
n:tells us how its moving radially relative to the nuc
ml,l: tells us how its moving angularly relative to the nuc.
orbital energies are negative meaning
the e- is bound to the nucleus
why does n=2 have 2 different wavefunction graphs
n=2 ,, l = 0,1 ml = -1,0,1
u get 2s and 2p
2 different wavefunctions bc u do radial x 2 different angular parts(different l values)
1 radial one is multiplied by 2 different angular parts to give 2 different products.
radial node:
where wavefunction graph passes through 0
angular node:
the origin of the graph ,, where it starts
orbitals with the same principal quantumn number =
same energy
they form a shell
orbitals with the same n and l values
same n so same shell
same l so same subshell
l=0=s l=1=p l=2=d etc
shell
same n value
subshell
different l value
same n number
2p and 2s
in a shell we should have how many subshells
u should have ‘n’ sub shells
how can we find out how many degenerate orbitals there are in a shell
n^2
tells us how many degenerate orbitals there are in a shell.
shell = same n value.
ground state §
1s
no n = 0 no no no
everything else is an excited state. but with different energy levels of the excited states.
are 2s and 2p degenerate
yes, for this lecture they are degenerate: for H atom!!!!
bc same shell : aka same ‘n’ number
l = different but l gives shape not energy
n^2
