3b

Question 1

schrödinger equationssss

Answer 1

HY=EY
(T’ + V’)Y = EY

Question 2

T’ =

Answer 2

(- h.dash/2m) X (2nd derivative) (y)

is the result an eigen function?? a possible solution for Y ??

Question 3

what’s potential energy,, V’ for a free particle

Answer 3

it’s 0

this is the model potential

Question 4

what’s a more realistic model

Answer 4

particle in a box

Question 5

describe the drawing for particle in a box

Answer 5

okay so u have a box in the middle with dashes outside the box.

outside the box,, where it’s dashed: Y and V are infinite.

inside the box: V=0 so H’=T’ (kinetic energy only)

the box prevents the e- from moving for infinity,, the e- can move within the box,, but cannot leave the box.

Question 6

when V = infinite,, what is Y

Answer 6

0

Question 7

sin,, cos and exp particle in a box explanation

Answer 7

box for each,, with dashes outside.
x axis: 0 -> L

sin(0)=0 so the graph starts at 0 and ends at 0,, it’s giving sad face,, rainbow that fits perfectly in the box

cos(0)=1 so graph starts on 1 on the y axis and goes down to -1 basically.

exp(0)=1 so graph starts on 1 and goes up up up. like a smiley face slanted to the left kinda.

sin is: eigen, finite, single, continuous and normalised as it starts and ends at 0.

cos and exp aren’t continuous bc they don’t start at 0,, they start at 1.

Question 8

what’s a boundary condition : particle in a box

Answer 8

it’s when when graph should reach 0 at the corners of the box graph.

the Y should = 0 at the ends of the box.

sin does this but cos and exp don’t. (cos goes down and exp goes up)

Question 9

bc one sin wave works and is both an eigen function, value and agrees with the born requirements,, will every sin wave do the same

Answer 9

nope!!
although they all start at 0,, they don’t all end at 0. not all of them respect the boundary condition.

Question 10

bc one sin wave works and is both an eigen function, value and agrees with the born requirements,, will every sin wave do the same

Answer 10

nope!!
although they all start at 0,, they don’t all end at 0. not all of them respect the boundary condition.

Question 11

quantisation of the sin graph,, bc every n(pi) is a node

aka 2n is a node,, so is 3n

Answer 11

sin(kL) = 0 (node)
kL=n pi ( n as in an integer)
K = n pi / L

n = 1,2,3,4 etc.

when sin(x) is plotted against x (//radians)

Question 12

in particle in a box,, how many directions can the e- move in

Answer 12

just one
left to right
is given: ‘x’ on the x axis

Question 13

spider in bath and particle in box explanation

Answer 13

aka the particle can move in the bath aka the box

but will never have enough energy to go over the bath walls,, where the potential energy,, V,, is infinite

Question 14

inside the box,, the V = what,, and what does that mean

Answer 14

V’ = 0

particle can be anywhere in the box,, just not outside it.

Question 15

start to end of the box

Answer 15

0 –> L

length of the box
V’ here is 0

Question 16

where V’ = infinite,, what does Y equal

Answer 16

Y = 0
probs of finding an e- here // outside the box = 0

Question 17

in the box // between 0–>L,, what does H’ equal

Answer 17

H’ = T’

bc H represents all the energies of that particle and usually = V’ + T’

so if v’ = 0,, only T is left

Question 18

what gies an eigenfunction when u take the 2nd derivative of it

Answer 18

sin
cos
exp

watch out for the constants tho

Question 19

why are cos and exp not continuous

Answer 19

bc theres a discontinuity: going from V=infinity and Y=0 to Y=1 in the box

bc cos(0) = 1 and exp(0) = 1 so on the ‘0’ on the x axis,, u will start the line on 1.

whereas for sin,, sin(0) = 0,, so u go from Y=0 outside the box to Y=0 inside the box

Question 20

what must be seen for a boundary condition to be satisfied

Answer 20

Y=0!!!!

end of graph//box must be where the line starts and ends.

Question 21

for the sin graph to end and start at Y=0,, what must sin(kL) equal to

and what does kL equal

and describe everything about this

Answer 21

it must equal to 0

kL=n pi
where n is an integer

so L must equal n pi/k

and thid only gives Y=0 when n = integer: 1,2,3,4

bc on a sin wave,, it touches the x axis at pi, 2pi, 3pi, 4pi etc

in radians

Question 22

different sin waves // waves in the particle in a box are different what

Answer 22

theyre different possible wavefunctions!

different arrangements (how the wave graph looks in particle in a box) have different energies.

aka if the waves look different,, theyre different possible wave functions,, and have different energies,, and different ‘n’ integer values.
in the kL=n pi equation

Question 23

so Yn (x) is equal to what (aka the wavefunction at a certain integer// different wavefunction) as a function of x

Answer 23

C sin(n pi x) // L

where C is a constant
so for different values of ‘n’ u get different wavefunctions

Question 24

for different values of ‘n’ ,, in the particle in a box model,, u get what

Answer 24

different va;ues of ‘n’ give u different wave functions

bc Yn(x) = C sin(n pi x) // L

Question 25

describe the sin wave in particle in a box where: n = 1

Answer 25

little hill

starts at 0 and ends at 0

so X=0 (start of box) and L= X
( end of box) bc these are the bottom corners of the box + theyre both 0

Question 26

describe the sin wave in the particle in a box where: n = 2

Answer 26

actual wave shape..
starts at X=0,, goes up,, goes down past the x axis,, then goes back up and reaches L=X=0

Question 27

describe the sin wave in particle in a box where n = 3

Answer 27

starts at x=0,, goes up,, does down, passes x axis,, goes back up past the x axis,, then goes back down and reaches X=L=0

Question 28

n= 3 and n=2 in terms of how the wave looks

and its properties

Answer 28

n = 3
3= number of half wavelengths
up, down, up down

n = 2
2 = number of half wavelengths
up, down, up

each value of n gives a different wave with the same amplitude but they have a different wavelength due to them having to squish more peaks//troughs in the same distance (the box x axis: x –> L)

Question 29

kinetic energy is like the second derivative,, what do we mean by that in terms of particle in a box

Answer 29

okay so for different values of n we get different possible wavefunctions

and each of these,, due to having a different ‘n’ value,, will have the same amplitude but different wavelengths and different amount of peaks.

due to the different amount of peaks // number of half waves,, we can see that some waves have a higher energy:

the more peaks + more movement = shorter wavelength// kinetic energy the wave has,, the higher the nergy of the wave

KE is changing more quickly!!

bc inside the box V’=0 so H’ = T’,, the only energy it has is Ke(T’),, so the more it moves, ,the more Ke it has,, meaning the more energy it has .

Question 30

sin wave is also used to describe what in organic chem

Answer 30

its used to describe molecular orbitals!!

esp for butadiene
there will be a node when the phases change.

all orbitals in phase= n=1 no node,, just the hill.

4 pi electrons (2 for each double bond) so 4 MO

Question 31

instead of Y,, what do we acc use and what does it mean

Answer 31

Y^2

its the probability of finding a particle at any position

Question 32

C ,, the normalised constant = what

so what is Y^2(x) equal to

and what shape does this give for n =1

Answer 32

2/L

= (2/L) (sin^2(n pi x /// L)

same shape but more like a graph and not a hill,, lower edges and higher peak.

inegrat of Y^2 dx with limits of 0 and L gives 1

Question 33

probablistic graph ,,, graph peak =

Answer 33

high probability of finding a particle there

Question 34

particle in a box,, what does the ‘n quantum number specify

Answer 34

the wavefunction and the energy.

Question 35

zeropoint energy for n is what value

Answer 35

n=1

n cannot be 0

Question 36

for n = 1 whats the E1 equation for an e-

Answer 36

E1 = h^2 // 8me L^2

Question 37

what is the main transition we would see a molecule do

Answer 37

the homo to lumo transition

Question 38

how do we know which orbital is the homo and which one is the lumo

Answer 38

count the number of pi electrons!!

then do an energy graph of energy being an arrow up

draw lines going up like a ladder and label them 1, 2, 3, 4, etc

each line obvs has 2e- so u fill them up with 8e- in total,, 2 in each line.

homo is the highest occupied one ,, lumo is the lowest unocupied one

this is the energy change // transition we would see.

Question 39

what is the denominator,, ‘L’

Answer 39

the length of the particle in ‘m’

so the larger the particle,, the greater the denominator bc L^2,, so a lower energy,, + lower energy transition between energy levels

smaller ‘L’ = smaller particle length = larger energy gap,, energy levels space out.

Question 40

what is tunnelling

Answer 40

a property that quantum particles have

quantum particles can tunnel through objects! think of the memes where ppl try to push through doors!!

Question 41

what is V(potential energy) when tunneling occurs

Answer 41

V = 0 except at the barrier

kinda opposite of particle in a box as there,, inside the box was v=0,, here,, outside the barrier = V=0

Question 42

when V=0 what can occur + where is V=0 seen on the tunneling diagram

Answer 42

the particle can move anywhere,, can go to infinity

kinetic energy = H

wavefunction = sin//cosine wave

Question 43

describe the tunneling experience

Answer 43

so ur a He atom,, and ur a quantum particle and therefore have a wavelength + wave properties: ur norally a sin//cosine wave

so ur waving around and can go anywhere bc V=0 until u reach a barrier,, at the barrier,, the V=is no longer 0. aka at the barrier,, ur energy is not only kinetic energy.

u hit the barrier and need to be continuous: so if u hit it u get dropped onto the x axis,, and this isnt continuous,, u arent a good wavefunction bc ur not continuous,, u kinda just dropped instead of being a wave.

what happens instead is that u experience exponential decay inside the barrier. so u go through the barrier as ur wavefunction extends through it ,, u kinda go down but slowly,, u dont get dropped bc u have a wavelike property. iu get through the barrier and keep on being a sin//cosine wave. ur amplitude decreases a bit tho due to exponential decay!! but u keep on being a wave. THIS IS TUNNELING!!

tunneling: exponential decay through a barrier where V is not 0,, and u pass through and remain being a sin//cosine graph but with a slightly lower amplitude due to the exponential decay.

Question 44

in the barrier: Y = what

Answer 44

Y = e ^ -kx

Question 45

what is k in Y = e ^ -kx ,, aka the waefunction in the barrier

Answer 45

its the tunnelling coefficient

k = squareroot( 2m(V-E) //h.dash)

Question 46

what does the extent of tunneling depend on: if smt is gonna tunnel through a barrier or not

Answer 46

barrier height
barrier width
mass of particle

Question 47

high barrier: can we tunnel through

Answer 47

low chance

bc Y = e^-kx

and k = root (2m(V-E)//h.dash)

Question 48

low barrier: can we tunnel

Answer 48

more likely to tunnel yessss

Question 49

narrow barrier: can we tunnel

Answer 49

yesss
bc less exponentisal decay,, it wont decay to 0

Question 50

wide barrier : can we tunnel

Answer 50

even if the barrier is short,, if its wide exponential decay can make the particle reach 0,, so harder to tunnel through.

think that exponential decay occurs for longer as u need to tunnel through a greater distance!!

Question 51

light particle: cancwe tunnel

Answer 51

yessss

less decay occurs

Question 52

heavy particles: can we tunnel

Answer 52

nope
more decay occurs due to the larger mass!!

cannot tunnel through.

Question 53

e- , protons, atoms , molecules : rate in terms of tunnelling abilities

Answer 53

e- : can tunnel a long way and through wide barriers

protons
atoms
molecules

due to them getting larger and larger.

Question 54

before tunnelling. and after tunneling the particle is aaaaa

Answer 54

a free particleeeee ayayayayya

bc V= 0

so its energy = due to kinetic enenrgy ,, how much it moves

H’ = T’