The Bohr Model of Atomic Structure: Study Guide (Physics 2 15.2)

What is inside an atom

Atoms have internal structure. The model you need has two parts:

• A nucleus at the centre. It is very small and positively charged. It contains protons (charge +e) and neutrons (no charge). Together, protons and neutrons are called nucleons.
• One or more electrons (charge −e) around the nucleus.

The evidence for a small, dense nucleus came from scattering experiments in Rutherford’s laboratory, which Rutherford explained with his nuclear model in 1911. Positively charged alpha particles were fired at thin gold foil. Most went straight through, but a very few bounced back sharply. Only a tiny, concentrated positive charge could push them back like that.

The sizes are very different. An atom is about 10⁻¹⁰ m across, but a nucleus is roughly 10⁻¹⁵ m to 10⁻¹⁴ m across. So most of an atom is empty space, with the electrons moving far from the nucleus.

Mass is in the nucleus

A proton and a neutron each have a mass of about 1.67 × 10⁻²⁷ kg. An electron has a mass of only 9.11 × 10⁻³¹ kg, about 1/1800 as much. So the mass of an atom is dominated by its protons and neutrons. The electrons add only a few hundredths of one percent.

Elements, isotopes and ions

• Element. Each element has its own number of protons, called the atomic number Z. Every carbon atom has 6 protons; every iron atom has 26. Change Z and you change the element.
• Isotope. Atoms of the same element can have different numbers of neutrons. The total number of protons and neutrons is the mass number A, and it identifies the isotope. Carbon-12 and carbon-14 are both carbon (Z = 6), with 6 and 8 neutrons. The number of neutrons is N = A − Z.
• Ion. A neutral atom has as many electrons as protons. An ion is an atom with a nonzero net charge because it has gained or lost electrons. Net charge = (protons − electrons) × e.

The number and arrangement of electrons decides how an atom interacts with other atoms: how it bonds and reacts. The nucleus decides which element it is and how heavy it is.

Nuclear notation

Nuclear notation writes the mass number at the top left of the chemical symbol and the atomic number at the bottom left. Any ionic charge goes at the top right.

The Bohr model

In 1913 Niels Bohr proposed a model of the hydrogen atom (one proton, one electron). It uses classical physics you already know, plus one new rule.

Classical part. The electron moves in a circle of radius r around the nucleus. The only force on it that matters is the electric attraction of the nucleus, given by Coulomb’s law. That force points to the centre, so it provides the centripetal force:

k|q₁q₂|/r² = mv²/r

For hydrogen, q₁ = +e and q₂ = −e, so ke²/r² = mₑv²/r. This gives the electron’s speed in an orbit of radius r:

v = √(ke²/(mₑr))

A smaller orbit means a stronger force and a faster electron.

The problem. Classical physics allows any radius, and so any energy. It also predicts that a charge moving in a circle should give out electromagnetic waves, lose energy and spiral into the nucleus. Real atoms are stable and have only certain energies (you will see the evidence from spectra in Topic 15.3). Bohr’s model had to add a rule that only certain orbits are allowed.

Why only some orbits: the standing-wave picture

Topic 15.1 gave a reason for Bohr’s rule. The electron has a de Broglie wavelength λ = h/(mₑv). Think of the electron’s wave wrapped around the orbit. As you go once round the circle, the wave must join up smoothly with itself. That is only possible if a whole number of wavelengths fits exactly into the circumference:

2πr = nλ, n = 1, 2, 3, …

This is the same idea as a string fixed at both ends, where only a whole number of half-wavelengths fits. If the wave does not fit, it does not join up, and that orbit is not allowed.

Putting the two conditions together

Each allowed orbit has to satisfy both conditions at once: the force condition (which links v and r) and the standing-wave condition (which links λ, and so v, to r). Only one radius satisfies both for each value of n. Combining them gives:

• r = n² r₁, where r₁ = 5.29 × 10⁻¹¹ m for hydrogen (n = 1, the smallest orbit);
• v = v₁/n, so the electron moves more slowly in larger orbits;
• λ = nλ₁, so the wavelength grows with n, and 2πr = n²(2πr₁) = n(nλ₁) still holds.

(If you substitute the constants, the radius comes out as h²/(4π²kmₑe²), which gives 5.3 × 10⁻¹¹ m with the rounded values in this course.)

Allowed orbits mean allowed energies

The energy of the electron-nucleus system is kinetic energy plus electric potential energy. Taking U = 0 when the electron is far away and at rest:

• K = ½mₑv² = ke²/(2r)
• U = −ke²/r
• E = K + U = −ke²/(2r)

The total energy is negative because the electron is bound: you must add energy to pull it free. Because only certain radii are allowed, only certain energies are allowed. For hydrogen these come out as E = −13.6 eV/n²: −13.6 eV for n = 1, −3.40 eV for n = 2, −1.51 eV for n = 3, and so on, getting closer to zero. These are the discrete energy states that Topic 15.3 draws on energy level diagrams.

What the Bohr model can and cannot do

The model works well for atoms or ions with a single electron (H, He⁺, Li²⁺). It does not work for atoms with several electrons, and electrons do not really follow sharp circular paths. In this course you describe electron structure only by energy levels. You do not need orbitals, orbital shapes or probability functions.

Worked example 1: counting particles in an ion

Question. An ion is written ⁵⁶₂₆Fe³⁺ (Figure 1). (a) How many protons, neutrons and electrons does it have? (b) What is its net charge in coulombs? (c) How does an atom of ⁵⁴₂₆Fe differ from a neutral atom of ⁵⁶₂₆Fe? (d) Estimate the fraction of the ion’s mass that comes from its electrons.

1. (a) Protons: Z = 26. Neutrons: N = A − Z = 56 − 26 = 30. The charge is 3+, so the ion has 3 fewer electrons than protons: 26 − 3 = 23 electrons.
2. (b) Net charge = (26 − 23)e = 3 × 1.60 × 10⁻¹⁹ C = +4.80 × 10⁻¹⁹ C.
3. (c) Same Z, so it is the same element with the same number of protons (26) and, if neutral, the same number of electrons (26). It has 54 − 26 = 28 neutrons, two fewer. It is a different isotope of iron. Its electron arrangement, and so its chemistry, is essentially the same.
4. (d) Nucleus: 56 × 1.67 × 10⁻²⁷ kg = 9.35 × 10⁻²⁶ kg. Electrons: 23 × 9.11 × 10⁻³¹ kg = 2.10 × 10⁻²⁹ kg. Fraction = 2.10 × 10⁻²⁹ ÷ 9.35 × 10⁻²⁶ ≈ 2.2 × 10⁻⁴, about 0.022%.

Interpretation. Losing three electrons changed the charge a lot but the mass by almost nothing. That is why ⁵⁶Fe³⁺ and a neutral ⁵⁶Fe atom have practically the same mass.

Worked example 2: the hydrogen ground state

Question. In the Bohr model, the electron in a hydrogen atom in its lowest state orbits at r₁ = 5.29 × 10⁻¹¹ m. Find (a) the electric force on the electron, (b) its speed, (c) its de Broglie wavelength, and check the standing-wave condition. (d) Find the total energy in eV.

1. (a) F = ke²/r² = (9.0 × 10⁹)(1.60 × 10⁻¹⁹)² ÷ (5.29 × 10⁻¹¹)² = 8.23 × 10⁻⁸ N, towards the nucleus.
2. (b) From ke²/r² = mₑv²/r: v = √(ke²/(mₑr)) = √[(9.0 × 10⁹)(1.60 × 10⁻¹⁹)² ÷ (9.11 × 10⁻³¹ × 5.29 × 10⁻¹¹)] = 2.19 × 10⁶ m/s. That is under 1% of c, so the non-relativistic formulas are fine.
3. (c) λ = h/(mₑv) = (6.63 × 10⁻³⁴) ÷ (9.11 × 10⁻³¹ × 2.187 × 10⁶) = 3.33 × 10⁻¹⁰ m.
4. Check: circumference 2πr₁ = 2π × 5.29 × 10⁻¹¹ m = 3.32 × 10⁻¹⁰ m. This equals λ (to within rounding of the constants), so exactly one wavelength fits: n = 1.
5. (d) E = −ke²/(2r₁) = −½ × (9.0 × 10⁹)(1.60 × 10⁻¹⁹)² ÷ (5.29 × 10⁻¹¹) = −2.18 × 10⁻¹⁸ J. Dividing by 1.60 × 10⁻¹⁹ J/eV: E = −13.6 eV.

Interpretation. The electron’s wavelength (3.3 × 10⁻¹⁰ m) is about the size of the atom itself. That is exactly the situation from Topic 15.1 where quantum theory is essential, so it makes sense that only certain states are allowed.

Worked example 3: comparing the n = 3 orbit

Question. Use the results of Worked example 2 to find, for the n = 3 orbit of hydrogen, the radius, speed, de Broglie wavelength and energy. Compare the electric force with the n = 1 value.

1. Radius: r₃ = 3² × r₁ = 9 × 5.29 × 10⁻¹¹ m = 4.76 × 10⁻¹⁰ m.
2. Speed: v ∝ 1/√r, and r is 9 times larger, so v₃ = v₁/3 = 7.29 × 10⁵ m/s.
3. Wavelength: λ ∝ 1/v, so λ₃ = 3λ₁ = 9.99 × 10⁻¹⁰ m.
4. Check: 2πr₃ = 2.99 × 10⁻⁹ m, and 2.99 × 10⁻⁹ ÷ 9.99 × 10⁻¹⁰ = 3.0. Three wavelengths fit, as n = 3 requires.
5. Energy: E ∝ 1/r, so E₃ = −13.6 eV ÷ 9 = −1.51 eV.
6. Force: F ∝ 1/r², so F₃ = F₁/81, about 1.2% of the ground-state force.

Interpretation. Higher n means a bigger, slower orbit with a weaker force and a higher (less negative) energy. Moving from n = 1 to n = 3 needs 13.6 − 1.51 = 12.1 eV of energy to be added to the atom. How that energy arrives, as a photon, is the subject of Topic 15.3.

Common misconceptions

• “The mass number is the mass of the atom in kilograms.” A is a count of nucleons. The mass is roughly A × 1.67 × 10⁻²⁷ kg.
• “Ions are different elements.” Gaining or losing electrons changes the charge, not Z. Fe³⁺ is still iron.
• “Isotopes have different numbers of protons.” Isotopes of an element have the same Z and different numbers of neutrons.
• “Electrons make up a good part of an atom’s mass.” They add only a few hundredths of a percent.
• “In a larger orbit the electron moves faster because it has further to go.” The force is weaker further out, so v = √(ke²/(mₑr)) is smaller.
• “An electron can orbit at any radius as long as the forces balance.” The force condition alone allows any radius. The standing-wave condition picks out only r = n²r₁.
• “The ground state has zero energy.” With zero set at infinite separation, the ground state has the most negative energy, −13.6 eV for hydrogen.
• Using q₁q₂ = e² for every atom. For a single-electron ion with Z protons the nucleus has charge +Ze, so the force is kZe²/r².

Where this leads

The allowed energies from the Bohr model explain why atoms emit and absorb light only at certain wavelengths. That is the next topic, emission and absorption spectra. If the de Broglie wavelength is unfamiliar, look back at quantum theory and wave-particle duality. Test yourself with the practice questions, then use the revision notes and the topic checklist to consolidate.