This unit is ≈4% of the A-Level Physics, across 8 lessons. Full syllabus
Lesson 8 of 8 · Options and board-specific extensions
Iterative modelling and the Boltzmann factor
7 min read · about 55 min with practice3 quick checks<1% of the testStretch: Stretch: harder material that separates the top grades
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OCR B (H557) expects you to build models, not just quote their solutions. Papers 1 and 2 ask you to run steps of an iterative model by hand, explain why it drifts from reality, and use the Boltzmann factor e−E/kT to explain why some processes switch on so sharply with temperature. Paper 3 wraps the same ideas in experimental data.
By the end you’ll be able to
Build and interpret iterative models of capacitor discharge and radioactive decay (ΔN = −λNΔt)
Model simple harmonic motion iteratively (a = −(k/m)x) and discuss step size and errors
Use the Boltzmann factor e^(−E/kT) to compare rates of thermally activated processes (evaporation, reaction rates, charge carriers)
Compare model predictions with experimental data and discuss the validity of assumptions
What the exam asks
Decay models: step ΔQ=−RCQΔt or Δ and compare with the exponential solution.
vii.Check your understanding
3 questions on iterative modelling and the Boltzmann factor. Every option is explained once you answer.
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PromptCard 1 of 3
Describe the basic loop of an iterative model.
N
=
−λNΔt
Oscillator models: from a=−mkx, update v and x in the order given; explain the effect of step size.
Exponential tools: time constant, half-life, the constant-ratio property and log-linear graphs.
Boltzmann factor: calculate e−E/kT, compare rates at two temperatures and explain activation processes.
Validity: compare model output with data and criticise the assumptions.
Core ideas
Why iterate?
When a rate of change depends on the quantity itself, predict it in small steps: treat the rate as constant over Δt, work out the change, update, and repeat. A spreadsheet does this thousands of times; the exam asks for two or three steps by hand.
Capacitor discharge and radioactive decay
The discharge current is I=RV=RCQ, so in one step
ΔQ=−RCQΔtQnew=Q(1−RCΔt)
Every step multiplies Q by the same factor, which is the discrete version of the constant-ratio property: in equal times, the quantity falls by equal fractions. The exact solution is
Q=Q0e−t/RCτ=RCT1/2=RCln2≈0.69RC
After one time constant, Q has fallen to 1/e≈37%. A graph of lnQ against t is a straight line with gradient −1/RC.
Radioactive decay is the same model with ΔN=−λNΔt and T1/2=λln2.
The simple model overestimates the decay. It uses the rate at the start of each step, which is the largest rate in that step, so too much charge leaves. The error shrinks as Δt becomes a smaller fraction of RC: keep Δt≲RC/100 for good accuracy.
An oscillator, step by step
For a mass on a spring, F=−kx, so a=−mkx. A typical loop is:
Find the acceleration from the current displacement: a=−mkx.
Update the velocity: vnew=v+aΔt (OCR B often writes this as momentum: Δp=FΔt).
Update the displacement: xnew=x+vnewΔt.
The model gives sinusoidal motion with period close to 2πm/kifΔt is small compared with the period (say T/100). With large steps the phase drifts and the amplitude can grow or shrink, so energy is not conserved. Adding −bv to the force models damping.
The Boltzmann factor
In a system at temperature T, energy is constantly shared out at random. The ratio of the numbers of particles in two states that differ in energy by E is
n1n2=e−E/kT
This is the Boltzmann factor. It is also, roughly, the fraction of particles that have at least energy E available at any instant. kT is the typical energy of thermal jostling (kT=0.025 eV at 290 K).
An activation process needs a particle to gain energy E before it can happen: a molecule escaping from a liquid, two molecules reacting, an electron breaking free to conduct in a semiconductor, a vacancy hopping in a metal. Its rate is roughly
rate∝(attempts per second)×e−E/kT
E/kT
e−E/kT
Typical result
1–5
0.37–0.007
happens almost at once
about 15–30
10−7–10−13
slow but noticeable, with a very strong dependence on T
over 60
below 10−26
effectively never happens
With about 1012–1013 attempts per second, a factor near 10−13 still lets each particle succeed about once every few seconds, which is why ordinary chemistry and evaporation work at room temperature. Because E/kT is large, a small rise in T produces a large rise in rate. Taking logs, ln(rate)=constant−kE⋅, so a graph of ln(rate) against 1/T is a straight line with gradient −E/k.
The atmosphere: raising a molecule of mass m through a height h costs E=mgh. In an isothermal atmosphere the number density therefore falls as n=n0e−mgh/kT.
Worked examples
Exam technique
Lay iterations out as a table (t, Q or x, v, a, change) and keep an extra significant figure until the end.
Convert eV to joules before dividing by kT, and use kelvin.
For ratios, subtract exponents:e−E/kT1. This avoids rounding errors.
Validity answers name a specific assumption (constant R or T, no damping, isothermal air), its effect on the prediction, and how the data show it.
Common mistakes
Quick recap
Iteration: rate from the current value → change = rate ×Δt → update → repeat.
Decay: Qnew=Q(1−Δt/RC); exact Q=Q0e−t/RC; T1/2=RCln2; lnQ against t has gradient −1/RC.
The simple model decays too fast; smaller Δt improves it.
Oscillator: a=−(k/m)x, then v, then x, in the order given; use Δt≪.
Boltzmann factor e−E/kT: population ratio for an energy gap E; rate ∝e.
E/kT of 15–30 gives slow but strongly temperature-dependent processes; ln(rate) against 1/T has gradient −E/k.