Quantum mechanics

Question 1

What is the Schrodinger equation in one dimension?

Answer 1

Question 2

What is the Schrodinger equation in three dimensions?

Answer 2

Question 3

What is the probability density?

Answer 3

In one dimension, the probability is proportional to ψ²(x)Δx, where ψ²(x) is the probability density. If Δx is small, the probability is ψ²(x)dx. In three dimensions, Ψ²(x,y,z)ΔxΔyΔz=ψ²(x,y,z)ΔV. If ΔV is very small, Ψ2(x,y,z)dxdydz=ψ2(x,y,z)dV.

Question 4

How do you normalise the wavefunction?

Answer 4

If ψ is a solution to the Schrodinger equation, so is Nψ, where N is any constant. It becomes the normalisation constant.

Question 5

What are acceptable wavefunctions?

Answer 5

They must be single-valued at any point. They must not be infinite over a measurable interval. They must be a continuous and smooth function. Their first derivative must be a continuous and smooth function. They must be quadratically integrable so the integral must be finite.

Question 6

What is the Heisengberg uncertainty principle?

Answer 6

It is not possible to specifiy simultaneously, with arbitrary precision, both the momentum and the position of the particle.

Question 7

What is the Schrodinger equation for a particle in a one-dimensional box?

Answer 7

Question 8

How do you find the solution for the Schrodinger equation for a particle in a one-dimensional box?

Answer 8

Question 9

What is the effect of the boundary conditions on the Schrodinger equation for a particle in a one-dimensional box?

Answer 9

Question 10

How do you normalise the wavefunction to calculate the constant A?

Answer 10

Question 11

What is the energy equation for a particle in a one-dimensional box?

Answer 11

Question 12

Describe a quantum particle in a one-dimensional box.

Answer 12

The kinetic energy is quantised. The lowest energy level (known as the zero-point energy) is h²/(8mL²). This agrees with the Heisenberg uncertainty principle. For n=1, the maximum probability density ψ² is at x = L/2; for n=2 the maxima are at x = L/4 and x = 3L/4. A point at which ψ=ψ²=0 is called a node. Generally, the number of nodes increases with the increase in energy. The energy n²h²/(8mL²) decreases with the increase of the mass of the particle m and the size of the box L. The effects of quantization become less important, the energy gradually becomes continuous and the particle begins to behave as expected in classical mechanics.

Question 13

Describe a classical particle in a one-dimensoinal box.

Answer 13

The kinetic energy varies continuously and can take any value. The lowest value that the energy can take is zero. This corresponds to a stationary particle. The probability of finding a moving particle anywhere in the box would be same irrespective of the energy. No differences for large m and/or large L.

Question 14

Describe the free electron model.

Answer 14

One approximate way of treating the π electrons in the double bonds of a polyene is to assume that they move along the conjugated chain as particles in a one-dimensional box. The potential energy inside the box (along the chain) is constant but rises sharply at the ends.

Question 15

What is tunnelling?

Answer 15

It is observed when the barrier is not infinite.

Question 16

How do you calculate the probability of a particle tunnelling through the barrier?

Answer 16

Question 17

What is the Schrodinger equation for a particle in a three-dimensional box?

Answer 17

Question 18

What is the solution for the Schrodinger equation of a particle in a three-dimensional box?

Answer 18

Question 19

What is the energy equation for a particle in a three-dimensional box?

Answer 19

Question 20

If the three-dimensional box is a cube, what is the wavefunction and energy of the particle?

Answer 20

Question 21

What is the Schrodinger equation for a particle in a ring?

Answer 21

Question 22

Give the probability density, normalisation integral and overlap integral of real and complex wavefunctions.

Answer 22

Question 23

What is the energy equation for a particle in a ring?

Answer 23

Question 24

Apply the boundary conditions to the wavefunction of a particle in a ring and normalise it.

Answer 24

The boundary condition is that after a complete turn around the ring, the wavefunction should remain unchanged.

Question 25

What is the equation for angular momentum?

Answer 25

m = mass

v = velocity

r = radius

p = linear momentum

Question 26

How do you calculate angular momentum from kinetic energy and total energy?

Answer 26

Question 27

What is the probability density of a particle in a ring?

Answer 27

The probability density is independent of the angle φ. The momentum and angular momentum of a particle in a ring have exact values, but the position is completely undefined.

Question 28

What is Hook’s law?

Answer 28

Question 29

What is the equation for the potential energy of a compressed or extended spring?

Answer 29

Question 30

What are the allowed energies of the quantum harmonic oscillator?

Answer 30

Question 31

How do you calculate the frequency and centre of mass of a quantum harmonic oscillator, similar to a diatomic molecule?

Answer 31

Question 32

What is an eigenvalue equation?

Answer 32

Question 33

What is an observable?

Answer 33

A measurable property of a quantum system.

Question 34

What is a Hermitian operator?

Answer 34

They only have real eigenvalues, even if the eigenfunctions are complex.

Question 35

What is the relationship between the Hamiltonian operator and the classical energy expression?

Answer 35

Question 36

What is the difference between the time-dependent and time-independent Schrodinger equations?

Answer 36

Question 37

What is the equation for the Coulombic potential energy?

Answer 37

Question 38

What is the Schrodinger equation for an electron in a hydrogenic atom with a fixed nucleus?

Answer 38

Question 39

What is the Schrodinger equation for an electron in a hydrogenic atom where the nucleus is free to move?

Answer 39

Question 40

What are the polar spherical coordinates?

Answer 40

Question 41

What is the solution for the Schrodinger equation of a hydrogenic atom using polar spherical coordinates?

Answer 41

Question 42

Which quantum numbers do the functions of the wavefunction, using polar spherical coordinates, depend on?

Answer 42

r depends on the principal quantum number (n) and the orbital angular momentum quantum number (l). θ depends on l and the magnetic quantum number (m_L). φ depends on m_L.

Question 43

How do you calculate the energy of a hydrogenic atom?

Answer 43

Question 44

What is the radial distribution function?

Answer 44

The radial distribution function r²R² when multiplied by dr gives the probability of finding the electron in a spherical shell with radius r and thickness dr centred on the nucleus. The volume of this shell is 4πr²dr, hence the probability of finding the electron at the nucleus is 0.

Question 45

Describe the p-orbitals.

Answer 45

p₀ is real and corresponds to the p_z orbital. p₊₁ and p_–1 are complex functions. As p₊₁ and p_–1 are degenerate (equal in energy), so can be combined to give the p_x and p_y orbitals. The value of the magnetic quantum number ml is not defined for the p_x and p_y orbitals. The 2p orbitals have a double-lobed appearance. The two lobes are separated by a nodal plane that cuts through the nucleus. The p orbitals for n > 2 have additional nodal surfaces due to their radial wavefunctions.

Question 46

Describe the d-orbitals.

Answer 46

The angular components of the d orbitals for different values of n stay the same, and the only real one is that for d₀ which corresponds to dz². Other real d orbitals can be obtained through suitable combinations of the complex ones.

Question 47

What is electron spin?

Answer 47

The spin of the electron is an intrinsic property of the electron which cannot be changed or eliminated. Electron spin is described by two quantum numbers:

• spin quantum number s = ½ 
• spin magnetic quantum number ms = +½,–½

Question 48

What are the alpha and beta spin functions?

Answer 48

It is convenient to think that α electrons and β electrons are described by the so-called α and β spin functions, α(ω) and β(ω), ω = +½,–½.

Question 49

What is the Schrodinger equation of the helium atom?

Answer 49

Helium has 2 electrons and a fixed nucleus.

H(1,2) = T(1) + T(2) + V(1) + V(2) + V(1,2)

Question 50

What is the orbital approximation?

Answer 50

Each electron has its ‘own’ spin orbital (the spin function σ can be α or β). One spatial orbital can be used twice, once with an α spin function and once with a β spin function. For helium, this spatial orbital is ‘doubly-occupied’, that is, it accommodates two electrons, one with spin α and a second one with spin β.

Question 51

What is the Pauli Exclusion principle?

Answer 51

A spin orbital may be occupied by a single electron only. No more than two electrons may occupy any given spatial orbital, and if two electrons do occupy the same spatial orbital, then their spins must be opposite. The many-electron wavefunction must be antisymmetric. In other words, it has to change sign when we exchange the positions of two electrons. The wavefunction for helium is not antisymmetric, however it can be made antisymmetric:

Ψ_A(1,2) = Ψ(1,2) - Ψ(2,1) =

ψ₁(1)α(1)ψ₁(2)β(2) - ψ₁(2)α(2)ψ₁(2)β(2)

This can be written as a Slater Determinant. If 2 electrons are placed in the same spin orbital, an antisymmetric wavefunction becomes equal to zero.

Question 52

If the exact wavefunction of the orbital approximation was available, how would you calculate its energy?

Answer 52

The resultant expression is called the energy expectation value corresponding to the wavefunction Ψ. It can be evaluated for both exact and approximate wavefunctions. dτ is the product of all variables included in Ψ(1,2,…,N) and the integration is over all allowed values of these variables.

Question 53

What is the variation principle?

Answer 53

According to the variation principle, the energy of an approximate wavefunction is always higher than that of the exact solution of the Schrödinger equation.

Question 54

What is penetration and shielding?

Answer 54

An electron at a distance r from the nucleus in a many-electron atom is shielded and experiences a reduced nuclear charge Z_eff. The s orbitals penetrate closer to the nucleus and experience less shielding than p orbitals. As a consequence of penetration and shielding, the energies of the orbitals in the same shell of a many-electron atom are ordered as s < p < d < f.

Question 55

What are valence electrons?

Answer 55

Valence electrons are the electrons in the outermost shell of an atom in its ground state. The aufbau (building-up) principle:

1. The order of occupation of atomic orbitals is: 1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s 5d 4f 6p …
2. According to the Pauli exclusion principle, each spatial atomic orbital can accommodate up to two electrons.
3. Electrons occupy different orbitals of a given subshell before doubly occupying any one of them.
4. In its ground state, an atom adopts a configuration with the greatest number of unpaired electrons (Hund’s rule).

Question 56

What is the Born-Oppenheimer approximation?

Answer 56

The electrons and nuclei in a molecule are assumed to move independently. The fast-moving electrons move around stationary nuclei, while the slow-moving nuclei experience the averaged influence of the fast-moving electrons. The Born-Oppenheimer approximation allows separation of the Schrödinger equation for a molecule into two equations, one for the electrons, and one for the nuclei. The electronic Schrödinger equation can be solved for many different positions of the fixed nuclei. For a diatomic molecule this produces a potential energy curve which shows how the molecular energy depends on the internuclear separation. For molecules with more than two atoms, it is possible to obtain a potential energy surface.

Question 57

What is the Schrodinger equation for the H₂ molecule?

Answer 57

The H₂ molecule has two electrons and two nuclei.

H(1,2) = T(1) + T(2) + V(1) + V(2) + V(1,2) + V(A,B)

Question 58

What is the linear combination of atomic orbitals?

Answer 58

The molecular orbital of H₂ can be expressed as:

ψ₁ = c_Aψ_A1s + c_Bψ_B1s

If we assume the MO is real, the probability density is:

ψ₁²= (c_Aψ_A1s + c_Bψ_B1s)² =

c_A²ψ_A1s² + c_B²ψ_B1s² + 2c_Ac_Bψ_A1sψ_B1s

The probability that the electron is mostly in orbital ψ_A should be just the same as the probability that the electron is mostly in orbital ψ_B, therefore c_A² = c_B² so c_A = +/- c_B

Therefore, c_A and c_B can combine to give N.

ψ₁ = N₁(ψ_A1s + ψ_B1s) for a bonding MO

ψ₂ = N₂(ψ_A1s - ψ_B1s) for an anti-bonding MO

ψ₁² shows that there is more electron density between the two nuclei. ψ₂² shows that there is no electron density halfway between the two nuclei.

MOs that do not change sign upon reflection in any plane passing through all nuclei are called σ orbitals. ψ₁ is the lowest-energy bonding σ orbital, therefore it is often labelled as 1σ. ψ₂ is the lowest-energy antibonding σ orbital, therefore it is often labelled as 1σ*.

Question 59

What are the changes in the energies and shapes of the H₂ MOs as the nuclei become well separated?

Answer 59

When the hydrogen atoms are well-apart, the MOs are almost degenerate, and their energies approach those of the H 1s orbitals.

Question 60

What are the changes in the energies and shapes of the H₂ MOs at the equilibrium H–H bond length?

Answer 60

Question 61

What are the changes in the energies and shapes of the H₂ MOs when the distance between the atoms is zero?

Answer 61

The nuclei merge and the 1σ_g MO transforms into a 1s AO, while the 1σ_u MO transforms into a 2p_z AO.