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E.1 Structure of the atom · HL only · Part 3 of 3 · student investigation · 80 minutes

Inside the nucleus and the Bohr atom (HL)

Predict how the distance of closest approach changes when the alpha particles get twice the energy, how much bigger a nucleus with 8 times the nucleons is, what goes wrong with Rutherford’s formula at high energy, and how big the n = 3 orbit of hydrogen is. Then test each prediction with a simulation.

1 Predict · 7
2 Closest approach · 16
3 Radius and density · 16
4 Deviations · 15
5 Bohr · 16
6 Show + reflect · 10

Learning goals: by the end you can…

find the distance of closest approach with energy conservation: E_k = kq₁q₂/d.
use R = R₀A^⅓ and show that every nucleus has the same density.
explain the deviations from Rutherford scattering with the strong nuclear force, and use E₀ to estimate a nuclear radius.
explain why Bohr introduced mvr = nh/2π and use r ∝ n², v ∝ 1/n and E = −13.6/n² eV (hydrogen only).
find the levels of a hydrogen-like ion from E = −K/n².

How to work in this lesson

How to work. Pairs discuss and agree, but each of you writes. Pairs → individual discuss the method, then write your own answer. Individual no discussion until your teacher says so. Your typed answers are saved on this device only.

Everything you need is on this page. Each simulation uses the same numbers as the questions next to it, and stays locked until you have made your prediction. Type powers of ten like 1.9×10^24 or 1.9e24. Give answers in the unit shown next to the box.

📘 From the data booklet

E_p = kq₁q₂/r (D.2)R = R₀A^⅓mvr = nh/2πE = −13.6/n² eV

Constants: k = 8.99 × 10⁹ N m² C⁻², e = 1.60 × 10⁻¹⁹ C, R₀ = 1.2 × 10⁻¹⁵ m, u = 1.66 × 10⁻²⁷ kg, h = 6.63 × 10⁻³⁴ J s, m_e = 9.11 × 10⁻³¹ kg. E = −13.6/n² eV is for hydrogen only. 1 MeV = 1.60 × 10⁻¹³ J.

2. Closest approach: test prediction 1.1

16 minPairs → individual

ATL Thinking: use conservation of energy. Self-management: charges (2e, Ze) and units (MeV to J).

Simulation A · Head-on to a nucleus the kinetic energy far away becomes electric potential energy; the particle stops at d

kinetic energy E_k 6.0 MeV

distance of closest approach d—

d in fm—

radius of the nucleus—

Distance of closest approach (head-on, nucleus fixed) E_k = kq₁q₂/dd = k(2e)(Ze)/E_k for an alpha particle; E_k in joules (MeV × 1.60 × 10⁻¹³). Use the proton number Z, not A.

2.1Alpha particle, 6.0 MeV, gold (Z = 79). d =

2.2Set E_k = 12.0 MeV. d =

2.3The gold-197 target is replaced by gold-195 (2 fewer neutrons). The distance of closest approach…

stays the same: Z is the same and neutrons have no charge gets smaller gets bigger

2.4Choose tin (Z = 50) and a proton. A proton accelerated from rest through 4.0 MV is fired head-on. d =

Back check

2.bBack check: an alpha particle accelerated through the same 4.0 MV, fired at tin. d =

Energy conservation, step by step

Far away: all kinetic energy, E_k; no potential energy.
At the distance of closest approach the particle stops for an instant: all electric potential energy, kq₁q₂/d.
Set them equal: E_k = kq₁q₂/d, so d = kq₁q₂/E_k. Change MeV to J.

Find the mistake

A student finds d for a 6.0 MeV alpha particle on gold-197:

d = k(e)(197e)/(6.0 × 1.60 × 10⁻¹³) = 4.7 × 10⁻¹⁴ m

What is wrong?

Two mistakes: the alpha charge is 2e, and the nuclear charge uses Z = 79, not A = 197. d = k(2e)(79e)/E_k = 3.8 × 10⁻¹⁴ m. The energy should stay in MeV. Nothing is wrong.

Hinge question

Choose your answer. You will see at once if it is right, and why.

At the same alpha energy, the target nucleus is changed to one with twice the proton number. The distance of closest approach…

A. halves B. doubles C. stays the same

2.5Was prediction 1.1 right? Explain with energy conservation.

What a good answer looks like

A, it halves. At closest approach all the kinetic energy has become electric potential energy: E_k = kq₁q₂/d, so d = kq₁q₂/E_k and d ∝ 1/E_k. With twice the energy the alpha particle can get twice as close: in the simulation 3.8 × 10⁻¹⁴ m at 6.0 MeV became 1.9 × 10⁻¹⁴ m at 12.0 MeV.

3. Radius and density: test prediction 1.2

16 minPairs → individual

ATL Thinking: show that a quantity cancels. Research: compare models with data.

Simulation B · How big is a nucleus? change the nucleon number A

nucleon number A 27

radius R = R₀A^⅓—

volume—

density—

Nuclear radius R = R₀A^⅓ R₀ = 1.2 × 10⁻¹⁵ m. Mass ≈ A × u (u = 1.66 × 10⁻²⁷ kg); volume = (4/3)πR³ = (4/3)πR₀³A, so A cancels: ρ = 3u/(4πR₀³) the same for every nucleus.

3.1Radius of gold-197, R =

3.2Radius of helium-4 (an alpha particle), R =

3.3Radius of uranium-238 ÷ radius of aluminium-27 =

3.4Move the slider: the density does not change. Its value, ρ =

kg m⁻³

What if? · stretch

3.wWhat would a teaspoon (5.0 cm³) of nuclear matter weigh? Its mass, m =

Find the mistake

A student compares two nuclei:

“Uranium-238 has about 9 times the nucleons of aluminium-27, so its radius is about 9 times bigger.”

What is wrong?

R ∝ A^⅓, not A: the radius is (238/27)^⅓ = 2.1 times bigger. It is the VOLUME that is about 9 times bigger. The radius is 9 times smaller. Nothing is wrong.

Hinge question

Choose your answer. You will see at once if it is right, and why.

Compare the density of a uranium nucleus with the density of a helium nucleus.

A. about the same B. uranium is about 60 times denser C. helium is denser

“Show that all nuclei have the same density”

Mass ≈ A u.
Volume = (4/3)πR³ = (4/3)π(R₀A^⅓)³ = (4/3)πR₀³A.
ρ = Au/((4/3)πR₀³A) = 3u/(4πR₀³): A cancels, so ρ does not depend on the nucleus.

3.5Was prediction 1.2 right? Explain with R = R₀A^⅓.

What a good answer looks like

B, 2 times. R ∝ A^⅓, and 8^⅓ = 2, so 8 times the nucleons gives twice the radius (and 8 times the volume). In the simulation helium-4 has R = 1.9 × 10⁻¹⁵ m and A = 32 has 1.2 × 10⁻¹⁵ × 32^⅓ = 3.8 × 10⁻¹⁵ m.

4. Deviations: test prediction 1.3

15 minPairs → individual

ATL Thinking: use a failed prediction as evidence. Communication: name the force and its range.

Simulation C · When does Rutherford’s formula fail? alpha particles scattered back from gold or aluminium; the graph shows measured ÷ predicted (Rutherford, electric force only)

alpha energy E 10 MeV

closest approach d—

touching distance R + R_α—

measured ÷ predicted—

Deviations: above an energy E₀ the alpha particle comes within the range of the strong nuclear force (about 10⁻¹⁵ m), so a second force acts and Rutherford’s formula (electric force only) no longer works. At E₀: R ≈ d = k(2e)(Ze)/E₀

4.1Gold: the deviations start at about 26 MeV. Distance of closest approach at 26 MeV, d =

4.2Use R = R₀A^⅓: R(gold-197) + R(helium-4) =

4.3Above E₀ the measured number falls below the prediction. Why?

the alpha particle comes within range of the strong nuclear force, so the electric force is not the only force the electric force stops acting the gold nucleus breaks into pieces at once

4.4Choose aluminium (Z = 13). Compared with gold, the deviations start at…

a lower energy a higher energy the same energy

What if? · stretch

4.wEstimate E₀ for aluminium-27, in MeV =

MeV

Find the mistake

A student explains the deviations:

“At high energy the electric force stops working, so the alpha particles go straight through.”

What is wrong?

The electric force still acts. A SECOND force, the strong nuclear force, also acts once the alpha particle is within about 10⁻¹⁵ m of the nucleus. The alpha particles are attracted by the electrons. Nothing is wrong.

“Outline why the deviation occurs” (1) · “how E₀ gives the nuclear radius” (2)

At closest approach the alpha particle is within the range of the strong nuclear force of the nucleus.
The nuclear radius is (approximately) the distance of closest approach at E₀.
From energy conservation: E₀ = kQq/r, so r = kQq/E₀.

4.5Was prediction 1.3 right? Explain with the distance of closest approach.

What a good answer looks like

A, the strong force. As the energy rises, d = k(2e)(79e)/E gets smaller. At about 26 MeV it is 8.7 × 10⁻¹⁵ m, about the distance at which the alpha particle and the gold nucleus touch (8.9 × 10⁻¹⁵ m). Closer than this, the short-range strong nuclear force acts as well as the electric force, so the electric-only formula of Rutherford no longer predicts the counts.

5. The Bohr model: test prediction 1.4

16 minIndividual

ATL Thinking: why a new rule was needed. Communication: the four marking points.

Simulation D · Bohr’s hydrogen atom choose an orbit; compare with the classical prediction

orbit n n = 1

radius r = n²r₁—

speed v—

energy E = −13.6/n²—

Bohr’s condition mvr = nh/2π n = 1, 2, 3 … Only orbits whose angular momentum is a whole-number multiple of h/2π are allowed, and in them the electron does not radiate. With ke²/r² = mv²/r: r = n²r₁E = −13.6/n² eV r₁ = 5.3 × 10⁻¹¹ m. Hydrogen only.

5.1Choose “classical”. Why did Bohr need a new rule?

an orbiting electron accelerates, so classically it radiates energy and spirals into the nucleus: atoms would be unstable the electron moves too slowly the nucleus attracts the electron too weakly

5.2Set n = 3. Orbit radius, r =

5.3Speed of the electron in n = 3, v =

m s⁻¹

5.4Energy of the electron in n = 3, E =

5.5A hydrogen-like ion (one electron) has levels E = −K/n². A photon of 40.8 eV moves the electron from n = 1 to n = 2. K =

What if? · stretch

5.wFor the same ion, the photon energy needed from n = 2 to n = 3 =

Find the mistake

A student finds the n = 2 energy of a helium ATOM (two electrons):

E = −13.6/2² = −3.40 eV

What is wrong?

E = −13.6/n² eV is for HYDROGEN only. For a one-electron ion the question gives the constant (E = −K/n²); for atoms with more electrons there is no formula: use the diagram or the data. It should be −13.6 × 2². Nothing is wrong.

Hinge question

Choose your answer. You will see at once if it is right, and why.

As n increases, the energy levels of hydrogen…

A. get closer together B. get further apart C. are equally spaced

5.6Was prediction 1.4 right? Explain with Bohr’s condition.

What a good answer looks like

B, 9 times. From ke²/r² = mv²/r and mvr = nh/2π, r = n² × h²/(4π²kme²) = n²r₁, so r ∝ n². For n = 3, r = 9r₁ = 4.8 × 10⁻¹⁰ m. The speed falls as 1/n (7.3 × 10⁵ m s⁻¹) and the energy rises to −1.51 eV.

6. Show it, then reflect

10 minIndividual

ATL Communication + Self-management: choose how to show your understanding, retrieve without notes, plan your next step.

Success criteria: your answer must…

use energy conservation with the right charges (2e, Ze);
use R = R₀A^⅓ correctly (cube root);
name the strong nuclear force and its short range;
use one number from a simulation today.

Option A · Written

One paragraph: “How scattering experiments measure the size of a nucleus.”

Option B · Sketch

The graph of measured ÷ predicted against energy for gold and aluminium, with E₀ marked and explained.

Option C · Spoken

Explain to your partner in 1 minute why Bohr introduced mvr = nh/2π. Your partner ticks the four marking points.

6.1Write your paragraph or your notes.

What a good answer looks like

Model answer (Option A): Alpha particles of increasing energy are fired at a target. At low energy the counts agree with Rutherford’s formula, because only the electric force acts. At the energy E₀ where the counts start to deviate, the alpha particle just reaches the nuclear surface and comes within range of the strong nuclear force. So the nuclear radius is about the distance of closest approach at E₀, found from energy conservation: E₀ = kQq/d. For gold E₀ ≈ 26 MeV gives d = 8.7 × 10⁻¹⁵ m, close to R(gold) + R(alpha) = 8.9 × 10⁻¹⁵ m from R = R₀A^⅓.

Exit ticket (no notes)

1An alpha particle of 5.0 MeV is fired head-on at gold (Z = 79). Distance of closest approach, in m =

2Radius of a nucleus with A = 64 (R₀ = 1.2 × 10⁻¹⁵ m), in m =

3Compared with a light nucleus, the density of a heavy nucleus is…

much bigger about the same much smaller

4Bohr radius r₁ = 5.3 × 10⁻¹¹ m. Radius of the n = 2 orbit of hydrogen, in m =

5The energy of the n = 4 level of hydrogen is…

−3.40 eV −0.85 eV −54.4 eV

6Which idea from today is still the least clear to you?

What a good answer looks like

There is no single right answer. A good answer names one idea exactly, for example: “I am not sure how the energy E₀ gives the radius of the nucleus.”

7Feedforward: Next time I will check ________ before I ________.

What a good answer looks like

A good feedforward names an action: “Next time I will check that I used Z (not A) and 2e for the alpha particle before I calculate d.”

TOK · 2 minutes

Theory of knowledge: how do we know? There is no single right answer. Write two or three sentences.

TOKBohr kept the electron in a classical orbit but simply forbade it to radiate, with no reason except that it worked. Is it acceptable for a scientist to add a rule only because it gives the right answer?

Points you could make

A rule that explains known results (the hydrogen lines) and predicts new ones (the lines in the ultraviolet and infrared) is useful even before it is understood.
It was accepted as a step, not a final answer: later quantum theory explained the rule with the wave nature of the electron.
The Bohr model also shows the limits of such rules: it fails for atoms with more than one electron.

Read the notes Part 1 Back to E.1