Equations

Question 1

E

Answer 1

Energy

Question 2

C

Answer 2

Speed of light (2.998x10^8m/s)

Question 3

ν

Answer 3

Frequency

Question 4

h

Answer 4

Plancks constant (6.626x10^-34 J•s)

Question 5

ħ

Answer 5

Reduced Planck’s constant (h/2π)

Question 6

P

Answer 6

P = kħ

Momentum

Question 7

F

Answer 7

F =ma
F=m((d^2x)/(dt^2))
(Force)

Question 8

P(x)=ψ*(x)•ψ(x)

Answer 8

ψ^2(x) = p(x)

probability distribution of finding the e-

Question 9

ψ(x)

Answer 9

Wave function
Particle
(wave particle duality)

Question 10

((-ħ^2)/(2m)•(d^2)/(dx^2))•ψ(x)

Answer 10

Kinetic energy

first half of left side of Schrödinger equation

Question 11

+ ν(x)•ψ(x)

Answer 11

Potential energy

Second half of left side Schrödinger equation

Question 12

= E ψ(x)

Answer 12

Total E of the e- (or the system)

Right side of Schrödinger equation

Question 13

Schrödinger equation

Answer 13

Solving gives us the wave function (x axis particle)

Squaring that gives us the probability of finding it

And the total energy of the system

(When 1d the potential energy = 0)

Question 14

ψ(x) types of waves

Answer 14

A•sin(k•x) = (ψ(x) the one we use in 1d Schrödinger equation generally)

B•cos(k•x) = (not 0 at edges ignore)

Ce^(ikx) = (advanced complex numbers ignore)

Question 15

Particle in box

Answer 15

Y axis goes to infinity it’s potential energy (which is v for some reason)

X axis goes from 0 to L less than 0 and right of L potential energy is infinite (so KE=0 I think)

Inside 1d box ν=0 so particle can only exist inside the box because ψ(x) must be continuous

Question 16

KE + PE = E(total)

Answer 16

Super simplified Schrödinger’s equation

Goes to E(total) = -KE (for some reason don’t need to learn)