Using Schrodingers Equation

Question 1

How do you derive the time independent Schrodinger equation?

Answer 1

You just add a term for the potential energy - TDSE + V(x) Ψ(x,t). Then substitute in Ψ(x,t) = ϕ(x)p(t), and rearrange

Question 2

What is the time independent Schrodinger equation?

Answer 2

-ħ^2/2m * d^2 ϕ/dx^2 + vϕ = Eϕ

Question 3

What is the equation for energy times momentum?

Answer 3

iħ * dp/dt = Ep

Question 4

What is the probability density given by for the TISE?

Answer 4

|Ψ|^2 = ϕϕ* pp* = |ϕ|^2

Question 5

What is a solution for the TISE?

Answer 5

p(t) = exp(-iEt/ħ)

Question 6

What is the infinite potential well problem and how is it solved?

Answer 6

A well between x=0 and x=L where the potential inside is 0 and outside in infinity. Need to solve the TISE by setting V(x) = 0, and k^2 = 2mE/ħ^2

Question 7

What is a solution to the TISE for the infinite potential well?

Answer 7

ϕ = Asin(kx) + Bcos(kx)

Substitute this into the TISE and see if it works.

Question 8

How do you find the constants for the solution of the TISE?

Answer 8

You find the constants using the boundary conditions (x=0 and x=L). To find A, need to use normalisation conditions - integrate A^2 sin^2 (npi*x/L) dx between 0 and L, and set the solution equal to 1. Rearrange to find A.

Question 9

What is the normalised wavefunction for a particle in the infinite potential well?

Answer 9

ϕ = (2/L)^(1/2)sin(npix/L)

Question 10

What are the energy levels for a particle in the infinite potential well?

Answer 10

En = (ħ^2k^2)/(2m) = (ħ^2n^2*pi^2)/(2mL^2)

Question 11

What is the finite potential well problem?

Answer 11

The same as the infinite potential well problem but with finite potential outside the well rather than infinite.

Question 12

How do you solve the finite potential well problem for inside the well?

Answer 12

Same as the infinite problem - d^2 ϕ/dx^2 + k^2 ϕ = 0, where k^2 = (2m*E)/(ħ^2)

Question 13

How do you solve the finite potential well problem for outside the well. Can we solve this?

Answer 13

Instead of setting V as 0,set it as V0. Then divide through by ħ/2m, and set equal to zero. Then set the constant before ϕ as α.

Question 14

How can we solve the finite potential well problem?

Answer 14

Have to match the wave functions inside and outside the well so that they satisfy the boundary conditions: Ψ(x) and dΨ(x)/dx must be continuous at x=0 and x=L.

Question 15

What is the finite potential barrier?

Answer 15

Finite potential well flipped upside down with 3 zones - before the barrier, inside the barrier and after the barrier.

Question 16

What are the TISE’s for regions 1 and 3 for the finite potential barrier?

Answer 16

1: d^2 ϕ1/dx^2 + (2mE)/(ħ^2) ϕ1 = 0
3: d^2 ϕ3/dx^2 + (2mE)/(ħ^2) ϕ3 = 0

Question 17

What are solutions for the equations for regions 1 and 3 in the finite potential barrier problem? What do these correspond to?

Answer 17

ϕ1 = Aexp(ikx) + Bexp(-ikx)
ϕ3 = Fexp(ikx) + Gexp(-ikx)
k = sqrt(2m*E)/ħ
The A term is the incident wave, the B term is the reflected wave, the F term is the transmitted wave and G is zero since there is no left travelling wave after the barrier.

Question 18

What is the transmission probability for the finite potential barrier?

Answer 18

(|ϕ3|^2)/(|ϕ1|^2) = FF/AA

Question 19

What is the TISE for region 2 in the finite potential barrier problem? What is a solution to this?

Answer 19

d^2 ϕ2/dx^2 + (2m)/(ħ^2) *(E-V)ϕ2 = 0

Solution is ϕ2 = Cexp(ik’x) + Dexp(-ik’x), with k’ = sqrt(2m(E-V))/ħ

Question 20

What can we say about the wavenumber of region 2?

Answer 20

It is imaginary because E is less than V. We can therefore replace -ik’ with k2.

Question 21

How do you find the constants of each region?

Answer 21

Use the boundary conditions where ϕ1=ϕ2 and ϕ2 = ϕ3: ik1A - ik1B = -k2C + k2D

and

-k2Cexp(-k2L) + k2Dexp(k2L) = ik3Fexp(ik3L)

Question 22

What is the A/F ratio?

Answer 22

A/F = (1/2 + (ik2/4k1))*exp((ik1+k2)L)

Question 23

What is the final equation for the transmission probability?

Answer 23

T ∝ exp(-2k2L) ∝ exp((-2Lsqrt(2m(V-E)))/ħ)

Question 24

What does the transmission probability equation mean?

Answer 24

That the wave function exponentially decreases within the barrier and then continues at the new amplitude after the barrier.

Question 25

What is scanning tunnelling microscopy?

Answer 25

A technique that allows conductive surfaces to be mapped at atomic resolution.

Question 26

What is the transmission probability of tunnelling phenomena?

Answer 26

T ~ exp(-(2L*sqrt(2m(V-E)))/ħ)

Question 27

What is an alpha particle?

Answer 27

2 protons and 2 neutrons bound together, with a nuclear binding energy of 28.3MeV.

Question 28

What is the Geiger-Nuttall law?

Answer 28

ln(λ) = -C1*((Z-2)/sqrt(E)) + C2, where λ is the decay constant, and Eis the energy of emitted alpha particles.

Question 29

What does the potential energy function for an alpha particle look like (interacting with a nucleus of radius R)?

Answer 29

• Potential energy on y axis and distance from centre of nucleus on x axis.
• Starts off below zero, then shoots up, then decreases by 1/r

Question 30

What is the equation for half life?

Answer 30

t(1/2) = ln(2)/λ = τ*ln(2), where τ is the reciprocal of the decay constant λ.

Question 31

What is the equation for number of nuclei remaining at time t (alpha decay)?

Answer 31

N(t) = N(0) exp(-λt)

Question 32

What is the approximate range of potential energies on the PE curve?

Answer 32

About 30 MeV.

Question 33

What is the equation for potential energy and kinetic energy of alpha particle?

Answer 33

Both the same equation - Ek = PE = (2(Z-2)e^2)/(4piϵ0d)