(a) (i) Show that ψ(x) = e^(αx) is an eigenfunction for the operator  = d^n/dx^n. What is the eigenvalue? (5 marks) (ii) The…

(a)(i) Eigenfunction of d^n/dx^n (5 marks)

Operate with  = dⁿ/dxⁿ on ψ(x) = e^(αx). Each differentiation brings down a factor α:

dψ/dx = α e^(αx), d²ψ/dx² = α² e^(αx), …, dⁿψ/dxⁿ = αⁿ e^(αx).

Thus Âψ = αⁿ ψ, which has the form Âψ = aψ with a constant a. Hence e^(αx) is an eigenfunction of dⁿ/dxⁿ, and the eigenvalue is αⁿ.

(a)(ii) Why n = 0 is rejected for the particle in a box (5 marks)

For n = 0, ψ₀(x) = √(2/L) sin(0) = 0 for every x. A wave function that is zero everywhere means there is no particle in the box, since the probability density |ψ|² is zero everywhere and the normalisation condition ∫₀^L |ψ|² dx = 1 cannot be satisfied. The state is therefore not physical.

The energy Eₙ = n²h²/8mL² also shows why: n = 0 would give zero energy, but a particle confined to a region of width L has an uncertainty in position Δx ≈ L, so by the uncertainty principle the momentum cannot be zero. The lowest allowed state, n = 1, has zero-point energy h²/8mL² > 0. Negative n gives the same states as positive n (only a change of sign).

(b)(i) Uncertainty in the momentum and speed of an electron (10 marks)

Using Δpₓ Δx ≥ h with Δx ≈ 50 pm = 5.0 × 10⁻¹¹ m:

Δpₓ ≥ h/Δx = 6.626 × 10⁻³⁴ J s/(5.0 × 10⁻¹¹ m) = 1.3 × 10⁻²³ kg m s⁻¹.

The corresponding minimum uncertainty in speed, with mₑ = 9.109 × 10⁻³¹ kg, is

Δv = Δpₓ/mₑ = 1.325 × 10⁻²³/9.109 × 10⁻³¹ = 1.5 × 10⁷ m s⁻¹.

Comments:

• The uncertainty in the speed is about 1.5 × 10⁷ m s⁻¹, which is about 5% of the speed of light and is roughly seven times the speed of the electron in the first Bohr orbit of hydrogen (2.2 × 10⁶ m s⁻¹). The momentum of an electron cannot be known precisely if it is confined to an atom-sized region, so it cannot be pictured as moving along a definite orbit with a known speed.
• This shows why the Bohr model with sharp orbits is not tenable and why the electron in an atom has to be described by a wave function and a probability distribution (an orbital). Such a large uncertainty would be entirely negligible for a macroscopic object, since Planck's constant is very small in comparison with its mass and size.

(b)(ii) Quantum numbers of a 2p_z orbital (5 marks)

For a 2p_z orbital, n = 2, l = 1 and m_l = 0.

• The principal quantum number n = 2 gives the shell (size and energy). The orbital has n − l − 1 = 0 radial nodes.
• The letter p corresponds to l = 1, the azimuthal (angular momentum) quantum number, which allows m_l = −1, 0, +1, three p orbitals. l = 1 gives one angular node, a plane.
• The subscript z shows the orientation along the z axis. The orbital with m_l = 0 has angular dependence cos θ, so it points along z with its nodal plane being the xy plane. The pₓ and p_y orbitals are combinations of the m_l = +1 and −1 functions.

(c)(i) Packing fraction of SC, BCC and FCC lattices (20 marks)

Packing fraction = (volume of the atoms in the unit cell, treating atoms as touching hard spheres)/(volume of the unit cell). Let the cell edge be a and the atomic radius r.

Simple cubic (SC). There are 8 × 1/8 = 1 atom per cell. The atoms touch along the cell edge, so 2r = a, r = a/2. Packing fraction = (4/3)π(a/2)³/a³ = π/6 = 0.524, or 52.4%. The coordination number is 6.

Body-centred cubic (BCC). There are 8 × 1/8 + 1 = 2 atoms per cell. The atoms touch along the body diagonal (length √3 a), which holds 4r, so r = √3 a/4. Packing fraction = 2 × (4/3)π(√3 a/4)³/a³ = π√3/8 = 0.680, or 68.0%. The coordination number is 8.

Face-centred cubic (FCC). There are 8 × 1/8 + 6 × 1/2 = 4 atoms per cell. The atoms touch along the face diagonal (length √2 a), which holds 4r, so r = √2 a/4 = a/(2√2). Packing fraction = 4 × (4/3)π(√2 a/4)³/a³ = π√2/6 = 0.740, or 74.0%. The coordination number is 12.

Conclusion: FCC is the most efficiently (most economically) packed, with 74% of the volume filled (the same as hcp, the maximum possible for equal spheres) and 26% empty space, while SC is the least, with 52.4%. BCC, at 68%, lies between them.

(c)(ii) Bragg's condition (5 marks)

Bragg's condition for constructive interference of X-rays scattered from a set of parallel lattice planes of spacing d is nλ = 2d sin θ, where λ is the wavelength of the X-rays, θ is the glancing angle between the incident beam and the planes, and n = 1, 2, 3, … is the order of reflection.

Explanation: rays reflected from successive planes travel different paths. The extra path length of the ray from the lower plane is 2d sin θ. The reflected rays are in phase, and reinforce each other, only when this path difference equals a whole number of wavelengths. At other angles the reflected rays are out of phase and cancel. Since sin θ ≤ 1, diffraction requires λ ≤ 2d, so the wavelength must be comparable with the interatomic spacing, as it is for X-rays (about 100 pm).