The structure of the atom IB DP Physics Study Notes - 2025 Syllabus

The structure of the atom IB DP Physics Study Notes

The structure of the atom IB DP Physics Study Notes at IITian Academy focus on specific topic and type of questions asked in actual exam. Study Notes focus on IB Physics syllabus with Students should understand

the Geiger–Marsden–Rutherford experiment and the discovery of the nucleus
nuclear notation _Z^AX where A is the nucleon number Z is the proton number and X is the chemical symbol
the relationship between the radius and the nucleon number for a nucleus as given by R = R₀A $^\frac{1}{3}$ and implications for nuclear densities
deviations from Rutherford scattering at high energies
the distance of closest approach in head-on scattering experiments

Standard level and higher level: 6 hours
Additional higher level: 3 hours

IB DP Physics 2025 -Study Notes -All Topics

Rutherford scattering

In 1897 British physicist J.J. Thomson discovered the electron, and went on to propose a “plum pudding” model of the atom in which all of the electrons were embedded in a spherical positive charge the size of the atom.
In the next slides we will disprove this model…
∙In 1911 British physicist Ernest Rutherford conducted experiments on the structure of the atom by sending alpha particles through gold leaf.
∙Gold leaf is like tin foil, but it can be made much thinner so that the alpha particles only travel through a thin layer of atoms.

FYI

An alpha (α) particle is a double-positively charged particle emitted by radioactive materials such as uranium.
Rutherford proposed that alpha particles would travel more or less straight through the atom without deflection if Thomson’s “plum pudding” model was correct:

FYI
∙Instead of observing minimal scattering as predicted by the “plum pudding” model, Rutherford observed the scattering as shown on the next slide:

Here we see that the deflections are much more scattered…

∙Rutherford proposed that all of the positive charge of the atom was located in the center, and he coined the term nucleus for this location.

FYI

∙IBO requires you to qualitatively understand the Geiger-Marsden scattering experiment.

∙Only by assuming a concentration of positive charge at the center of the atom, as opposed to “spread out” as in the plum pudding model, could Rutherford and his team explain the results of the experiment.

The nuclear radius

Now let’s calculate a ballpark figure for the nuclear radius by firing an alpha particle ($q = +2e$) at a nucleus ($Q = +Ze$). Assume the α begins far enough away that there is no $E_P$ between it and the nucleus.

$E_0 = E_{K0} + E_{P0} = E_{K\alpha}$

But as the α approaches the nucleus, repulsion will occur, and $E_P = \frac{kQq}{r}$ will increase, slowing it down.

In fact, at closest approach $R_0$, the α will momentarily stop before reversing direction.

Thus at the point of closest approach $E_K = 0$ and

$E = E_K + E_P = \frac{kQq}{R_0} = \frac{kZe(2e)}{R_0} = \frac{2Zke^2}{R_0}$.

∙Though its proof is beyond the scope of this course, the physical radius of the nucleus also depends on its neutrons, which contribute no charge. Thus the atomic mass number A is used, and here is the result:

Example:

Use the empirical formula for nuclear radius to calculate the radius of a gold nucleus ($ \ce{^{197}Au} $).

▶️ Answer/Explanation

The nuclear radius is given by the formula:

$ R = R_0 \cdot A^{1/3} $

Where:

$ R_0 \approx 1.2 \times 10^{-15} \, \text{m} $
$ A = 197 $ for gold-197

Substitute the values:

$ R = 1.2 \times 10^{-15} \cdot (197)^{1/3} $

$ R \approx 1.2 \times 10^{-15} \cdot 5.82 \approx 6.98 \times 10^{-15} \, \text{m} $

Isotopes

∙ Recall the mass spectrometer, in which an atom is stripped of its electrons and accelerated through a voltage into a magnetic field.
∙ Scientists discovered that hydrogen nuclei had three different masses.
∙ Since the charge of the hydrogen nucleus is $ e $, scientists postulated the existence of a neutral particle called the neutron, which added mass without charge.

∙ The proton and neutron are called nucleons.

∙ For the element hydrogen, it was found that its nucleus exists in three forms.

∙ A set of nuclei for a single element having different numbers of neutrons is called isotopes.
∙ A particular isotope of an element is called a species or a nuclide.
∙ An element’s chemistry is determined by the number of electrons surrounding it.
∙ The number of electrons an element has is determined by the number of protons in that element’s nucleus.
∙ Therefore, it follows that isotopes of an element have the same chemical properties.
∙ For example, there is water, made of hydrogen $ H $ and oxygen $ O $, with the molecular structure $ H_2O $.
∙ But there is also heavy water, made of deuterium $ D $ and oxygen $ O $, with the molecular formula $ D_2O $.
∙ Both have exactly the same chemical properties.
∙ However, heavy water is slightly denser than regular water.

∙ A species or nuclide of an element is described by three integers:

The nucleon number $ A $ is the total number of protons and neutrons in the nucleus.
The proton number $ Z $ is the number of protons in the nucleus. It is also known as the atomic number.
The neutron number $ N $ is the number of neutrons in the nucleus.

∙ It follows that the relationship between all three numbers is:

In nuclear physics, you need to be able to distinguish between different isotopes.

Example:

Carbon has two stable isotopes: $ \ce{^{12}C} $ and $ \ce{^{13}C} $. Compare the number of protons, neutrons, and nucleons in both isotopes.

▶️ Answer/Explanation

Step 1: Define key terms

Isotopes: Atoms of the same element with the same number of protons but different numbers of neutrons.
Nucleons: Protons + Neutrons in the nucleus.

Step 2: Analyze each isotope

$ \ce{^{12}C} $:
- Protons = 6 (atomic number)
- Neutrons = $12 – 6 = 6$
- Nucleons = $6 + 6 = 12$
$ \ce{^{13}C} $:
- Protons = 6
- Neutrons = $13 – 6 = 7$
- Nucleons = $6 + 7 = 13$

Conclusion:

The two isotopes have the same number of protons but differ by one neutron. Hence, the total number of nucleons is different: 12 for $ \ce{^{12}C} $, and 13 for $ \ce{^{13}C} $.

Fundamental Forces and Their Properties

∙ Given that a nucleus is roughly $ 10^{-15} \, \text{m} $ in diameter, it should be clear that the Coulomb repulsion between protons within the nucleus must be enormous.
∙ Since most nuclei do NOT spew out their protons, there must be a nucleon force that acts within the confines of the nucleus to overcome the Coulomb force.
∙ We call this nucleon force the strong force.

In summary, the strong force:
1. Counters the Coulomb force to prevent nuclear decay and therefore must be very strong.
2. Is very short-range, since protons located far enough apart do indeed repel.

Example:

Discuss the four fundamental forces of nature in terms of the following properties:

Relative strength
Range
Force carriers (mediating particles)
Particles affected

▶️ Answer/Explanation

Gravitational Force:
- Strength: Weakest (relative strength ≈ $10^{-39}$)
- Range: Infinite
- Carrier: Graviton (hypothetical)
- Affects: All particles with mass
Electromagnetic Force:
- Strength: Stronger than gravity (≈ $10^{-2}$)
- Range: Infinite
- Carrier: Photon
- Affects: Charged particles
Weak Nuclear Force:
- Strength: Weak (≈ $10^{-13}$)
- Range: Very short (≈ $10^{-18}$ m)
- Carriers: $W^+$, $W^-$, and $Z^0$ bosons
- Affects: All fermions (quarks and leptons); responsible for beta decay
Strong Nuclear Force:
- Strength: Strongest (relative strength = 1)
- Range: Very short (≈ $10^{-15}$ m)
- Carrier: Gluon
- Affects: Quarks (holds protons and neutrons together)

Question

What is the variation of nuclear density $\rho$ with nucleon number A ?

▶️Answer/Explanation

Ans:B

Question

The diameter of a nucleus of a particular nuclide $X$ is $12 \mathrm{fm}$. What is the nucleon number of $X$ ?
A. 5
B. 10
C. 125
D. 155

▶️Answer/Explanation

Ans:C

The structure of the atom IB DP Physics Study Notes - 2025 Syllabus