An introduction to modern physics

6 visual steps · 5 worked examples · 7 exercises with solutions

First, let’s see the idea.

Can more light compensate for insufficient energy per photon?

Move forward one step at a time. Change the values and observe what happens.

Each photon must be enoughSurfaceλ = 400 nm
Drawing description

A light wavefront with wavelength 400 nm travels towards the surface.

Step 1 of 6

Distinguish one photon from the amount of light

Change λ and observe the spacing of yellow ripples.

Light has a single wavelength λ. In this model each electron receives energy from one photon; reducing λ increases each photon's energy without requiring more photons.

E≈1239,84/λ(nm) eV
Wavelength
400 nm

What if you change this?

250 nm800 nm
1,5 eV4 eV
15

Before moving on

With λ=600 nm and φ=2,5 eV, does increasing flux from 1 to 5 cause emission?

Compare with the explanation

Not in this model: each photon has only about 2,07 eV, below 2,5 eV. Frequency must increase.

You decide when to mark this lesson as studied.

Each photon must be enough

Slide the drawing to see the details.

Explore the full explanationThe foundations, principles, worked examples and all the exercises, at your own pace.

What you will understand

  • Recognize the domains where classical models need extending.
  • Relate a photon's energy and frequency and interpret the photoelectric effect.
  • Connect atomic levels and spectra with energy differences.
  • Distinguish electronic and nuclear phenomena and interpret half-life as a statistical law.
  • Interpret rest energy and the relationship between mass, energy and speed in special relativity.

Before you start

You do not need to know everything by heart. The explanation revisits the necessary ideas and shows how they connect. If a foundation is still unclear, you can return to these lessons:

A model can work very well and still have limits

Newton's laws allow us to study object motion in many everyday situations. Wave optics explains interference and diffraction. The success of these models does not mean that each describes all phenomena at atomic scales or at speeds close to that of light. A model is evaluated by its predictions and the conditions under which it agrees with experiments.

When experiments reveal a regularity incompatible with a model, we need to review its assumptions. That does not turn all its results into errors. Classical mechanics remains an excellent approximation for many macroscopic motions at speeds much smaller than c. A broader theory must explain why that approximation works in its domain.

In this lesson we will build an initial connection between three areas: quantum physics, atomic/nuclear physics and special relativity. They are not a collection of mysterious names: they answer concrete questions. Why do certain metals emit electrons only above a particular light frequency? Why do atoms produce spectral lines? Why is the energy of a massive particle not given by mv²/2 when v approaches c?

Review energy, frequency and microscopic scales

Energy is measured in joules, J. For atomic processes, a joule is a very large unit compared with the energies involved. We often use the electronvolt, eV: the energy corresponding to the magnitude of the elementary charge crossing a potential difference of one volt. The conversion is approximately 1 eV = 1,602 × 10⁻¹⁹ J. The eV is energy, not voltage.

A wave's frequency f is the number of cycles per second, in Hz. In a vacuum, electromagnetic radiation obeys c = λf, where λ is wavelength and c ≈ 3,00 × 10⁸ m/s. The smaller λ, the greater f. Do not confuse frequency with intensity: a beam's total energy can be increased without changing frequency by increasing the amount of radiation crossing an area per unit time.

In calculations we will use h ≈ 6,626 × 10⁻³⁴ J·s for Planck's constant. A constant with such a small value in SI units helps explain why quantum effects are more evident at microscopic scales. Powers of ten require care: 1 nm = 10⁻⁹ m and 1 μm = 10⁻⁶ m.

What quantization of radiation means

Electromagnetic radiation can exchange energy with matter in quanta called photons. The energy of a photon with frequency f is E = hf, or E = hc/λ in a vacuum. This relationship connects frequency with the energy of an individual interaction. The product J·s × s⁻¹ has units of joules.

A photon with higher frequency has more energy. Increasing only the intensity of a fixed-frequency beam means, in the model, increasing the average number of photons arriving per unit time and area. It does not increase each photon's energy hf. If the beam carries total energy Ebeam with photons of the same frequency, the approximate number will be Ebeam/(hf).

This does not eliminate the wave description: interference and diffraction are still observed. Quantum physics predicts both the statistical pattern of many events and the localized detection of each event. Classical images of a “spread-out material wave” and a “little ball” are limited analogies. The quantum formalism does not require a particle to be literally a small sphere oscillating like a string.

The photoelectric effect: interpreting a threshold

Illuminating a suitable surface can emit electrons. In the simple one-photon-per-electron photoelectric description, some of the received energy is needed to free the electron from the material. This minimum energy is called the work function, φ, and depends on the surface. Energy conservation leads to Kmax = hf − φ, when hf is sufficient to allow emission.

The threshold frequency is f₀ = φ/h. Below it, the calculation hf − φ would be negative: that does not mean the electron escaped with negative kinetic energy. It means emission is not allowed in this model. Above the threshold, increasing frequency increases maximum kinetic energy; increasing intensity at fixed frequency can increase the number of emitted electrons, subject to surface and device conditions.

In an experiment with a retarding potential, the magnitude Ustopping needed to prevent the most energetic electrons from arriving satisfies eUstopping = Kmax. Not all emitted electrons have maximum energy, so this relationship describes one end of the distribution. The introductory explanation assumes suitable radiation and operating conditions; multiphoton phenomena and other details require more complete models.

Matter with wavelength and probabilistic predictions

Diffraction is not exclusive to light. Electrons and other material particles can also produce interference patterns. The de Broglie relationship associates wavelength with momentum: λ = h/p. For a particle of mass m moving sufficiently slowly compared with c, we can use p ≈ mv; the momentum relationship needs a relativistic correction at high speeds.

A macroscopic particle with large momentum in SI units has an extremely small wavelength. This helps explain why we do not normally see a ball diffract when passing through a doorway. An electron, however, can have a wavelength comparable to the separation between atoms in a crystal, allowing observable effects. Comparing scales matters as much as the formula.

The quantum state allows calculation of probabilities for measurement results. This is not merely a lack of experimental care: experiments prepared in the same way produce distributions predicted by the theory. The uncertainty principle for position and momentum can be written ΔxΔp ≥ h/(4π), where Δx and Δp represent appropriate statistical spreads. It does not claim that “nothing can be known”, nor is it merely an instrument malfunction; it limits simultaneous preparation of these distributions.

Energy levels and spectral lines

An atom contains a nucleus and electrons. Electrical attraction matters to atomic structure, but imagining electrons travelling in classical planetary orbits does not adequately describe all phenomena. In the quantum treatment, allowed electronic states have characteristic energies and spatial distributions; an orbital is not a rigid track along which an electron is observed circulating.

When moving from a higher-energy state Ei to a lower one Ef, the atom can emit a photon with hf = Ei − Ef. In absorption, an energy difference can be supplied by radiation. Because bound-state energies do not fill an arbitrary continuum, lines associated with certain differences are observed. Selection rules and other details also exist: matching energies is an important condition, not a complete description of all transition probabilities.

For hydrogen levels in an introductory model, En ≈ −13,6 eV/n², with n a positive integer and zero reference at the ionization threshold. The negative sign indicates a bound state. The state n = 1 is lower than n = 2. The formula predicts a basic structure of a one-electron atom well, but must not be used unchanged for all atoms. The modern quantum description also explains aspects that the Bohr model does not describe.

Diagram of three energy levels: a transition to a lower level emits a photonn = 1, lowest energyn = 2n = 3Photon E = E₃ − E₂Qualitative diagram: does not represent orbits.
Photon energy is the positive difference between the levels; it is not the negative value of one isolated level.

The nucleus and types of radioactive transformation

The nucleus contains protons and, in the usual isotopes except the simplest hydrogen, neutrons. The proton number Z identifies the element. Mass number A counts protons and neutrons; it is not mass in kilograms. Isotopes of the same element have the same Z and different neutron numbers. Electron transitions and nuclear transformations belong to different scales and mechanisms.

Some nuclei are unstable and transform spontaneously. In alpha emission, the nucleus emits a particle containing two protons and two neutrons: the residual nucleus has Z reduced by 2 and A by 4. In beta-minus emission, a neutron becomes a proton with emission of an electron and an antineutrino: Z increases by 1 and A does not change. The beta electron is not simply an electron that was already orbiting the nucleus and was pulled away. In gamma emission, the nucleus enters a lower-energy state by emitting a photon, without changing Z or A.

In reactions, the appropriate quantities of the complete system are conserved, including energy, momentum and charge. Ionization means removing or adding electrons to an atom; it is not the same as altering its nucleus. Ionizing radiation can produce ionization, but activity, energy per event and the effect in a material are different quantities. This lesson covers models without guiding experiments with radioactive sources.

A statistical law is not a clock for each nucleus

For a large population of nuclei of a given radionuclide in the exponential-decay model, the expected number not yet transformed is N(t) = N₀e−λt. The decay constant λ is positive and has units of inverse time. The negative sign describes the decrease of the original population. Transformed nuclei have not ceased to exist as matter: they have produced decay products.

The half-life T1/2 satisfies N(T1/2) = N₀/2. The exponential relationship gives λ = ln 2/T1/2 and N(t) = N₀(1/2)t/T₁/₂. After two half-lives, one quarter remains on average, not zero. The law does not choose which nucleus transforms first. For small populations, relative fluctuations matter.

Activity A = λN measures the expected number of transformations per second when λ is in s⁻¹; the unit is becquerel, Bq. Do not confuse activity with energy released per second, which also depends on each event's energy and the products. In decay chains, products may still be unstable; a single exponential may not describe all measured activity. Always specify which population and process were modeled.

Reference frames, the speed of light and rest energy

Special relativity starts from equivalence of physical laws in inertial frames and invariance of the speed of light in vacuum, c, for those observers. An inertial frame is one in which a body with no net force has uniform straight-line motion. The theory describes space and time measurements between observers; it does not claim all statements are relative or that any value can be chosen.

Comparing clocks and lengths involves defining events and simultaneity in each frame. One consequence is the factor γ = 1/√(1 − v²/c²). For a particle with rest mass m, its total free-particle energy is E = γmc², rest energy is E₀ = mc² and kinetic energy is K = (γ − 1)mc². When v is much smaller than c, the approximation recovers K ≈ mv²/2.

As v approaches c, γ grows without limit; a massive particle cannot be accelerated to c by a finite energy input. A photon has zero rest mass and energy E = pc: we must not substitute m = 0 into γmc² with v = c, since γ is not finite there. Formulas need their domains, just as in algebra.

What changes in a reaction that releases energy

A system's rest mass is linked to its energy in the frame where total momentum is zero. If a collection loses energy to the outside, the mass of the remaining system can decrease. In nuclear physics, a difference between total reactant and product rest masses can correspond to kinetic energy and radiation: reaction Q = (initial m − final m)c², correctly including all relevant particles.

It is not correct simply to say that “mass disappears” and forget the energy that left. The balance for the complete system still conserves energy and momentum. Nor can all of an object's mass be arbitrarily transformed into useful energy in a device: the fact that mc² is large does not describe an available mechanism or an efficiency.

These ideas connect topics already studied. A photon carries energy and momentum; a nucleus can emit a photon and recoil; a transition must respect the system's balance. Understanding the principles prepares later study of quantum and relativistic equations. Here we seek conceptual foundations and introductory calculations, without presenting a complete theory in a few pages.

Worked examples, without skipping steps

Follow the reasoning first. Then cover the solution and try to reconstruct it in your own words.

Example 1

From wavelength to emission threshold

Light of wavelength 600 nm in a vacuum strikes an idealized surface with work function 1,50 eV. Determine whether emission occurs in the one-photon-per-electron model and find maximum kinetic energy.
  1. We convert λ = 600 × 10⁻⁹ m and calculate E = hc/λ = (6,626 × 10⁻³⁴ × 3,00 × 10⁸)/(600 × 10⁻⁹) = 3,313 × 10⁻¹⁹ J.
  2. In eV, E ≈ 3,313/1,602 = 2,068 eV. Since 2,068 eV is greater than 1,50 eV, the model allows emission.
  3. Maximum K = 2,068 − 1,50 = 0,568 eV. The magnitude of the stopping potential would be approximately 0,568 V.
Result
Emission is possible; maximum K ≈ 0,57 eV.
Check the result
Maximum energy is positive and smaller than the photon's energy. A negative result would have indicated no emission in this model, not a physical negative kinetic energy.

Example 2

A hydrogen transition

In the model Eₙ = −13,6 eV/n², calculate the energy of the photon emitted in a transition from n = 3 to n = 2.
  1. E₃ = −13,6/9 ≈ −1,511 eV and E₂ = −13,6/4 = −3,40 eV. Level 2 has lower energy.
  2. The emitted energy is E₃ − E₂ ≈ 1,889 eV, positive. We do not add the magnitudes of the two levels.
  3. Converting to joules and using λ = hc/E gives approximately 6,57 × 10⁻⁷ m, or 657 nm with the rounded constants used.
Result
Photon energy ≈ 1,89 eV; wavelength approximately 657 nm.
Check the result
Rounding and the simplified model explain small differences from more precise spectroscopic values. The reverse transition would require absorption under appropriate conditions.

Example 3

Three half-lives do not eliminate the population

A sample initially contains 1,6 × 10⁹ nuclei of a radionuclide with half-life 20 h. Calculate the expected population after 60 h and the corresponding activity.
  1. 60/20 = 3 half-lives have elapsed. N = 1,6 × 10⁹/2³ = 2,0 × 10⁸ expected nuclei.
  2. We convert half-life to seconds: 20 × 3600 = 72000 s. λ = ln 2/72000 ≈ 9,63 × 10⁻⁶ s⁻¹.
  3. A = λN ≈ 9,63 × 10⁻⁶ × 2,0 × 10⁸ ≈ 1925 Bq. We consider only this radionuclide's population.
Result
Expected N = 2,0 × 10⁸; activity ≈ 1,93 × 10³ Bq.
Check the result
One eighth of the initial population remains, and this radionuclide's activity is also one eighth of the initial value. The prediction is statistical.

Example 4

A matter wavelength

An electron moves at speed 2,0 × 10⁶ m/s. Use m ≈ 9,11 × 10⁻³¹ kg and estimate its de Broglie wavelength.
  1. The ratio v/c is approximately 0,0067, so p ≈ mv is a suitable approximation for this estimate.
  2. p ≈ 9,11 × 10⁻³¹ × 2,0 × 10⁶ = 1,822 × 10⁻²⁴ kg·m/s.
  3. λ = h/p ≈ 6,626 × 10⁻³⁴ / 1,822 × 10⁻²⁴ = 3,64 × 10⁻¹⁰ m.
Result
λ ≈ 0,364 nm.
Check the result
J·s divided by kg·m/s gives metres. The scale is compatible with possible diffraction in atomic structures; it does not represent the electron's diameter.

Example 5

Kinetic energy at high speed

A particle has rest energy E₀ = 10 MeV and speed 0,60c. Determine total energy and kinetic energy.
  1. γ = 1/√(1 − 0,60²) = 1/√0,64 = 1,25. The factor is dimensionless.
  2. E = γE₀ = 1,25 × 10 = 12,5 MeV.
  3. K = E − E₀ = 2,5 MeV. Identifying total energy with kinetic energy is incorrect.
Result
E = 12,5 MeV and K = 2,5 MeV.
Check the result
If speed became zero, γ would be 1, E would be 10 MeV and K would be zero. This limit checks the meaning of rest energy.

Now try it yourself

There is no timer. Write what you know, choose a relationship and justify each step. Hints and solutions are available whenever you need them.

Exercise 1

Two beams have the same frequency, but the second has twice the intensity. Does each photon's energy double?
A hint
Separate individual energy hf from the average number of photons in the beam.
See the worked solution

No. Frequency f is the same, so hf is the same for every photon. Under the same geometric and duration conditions, the more intense beam carries more photons and more total energy on average. Intensity and frequency must not be confused.

Exercise 2

A surface has work function 2,4 eV and receives 2,0 eV photons. Calculate maximum K in the simple model, or explain why that physical value does not exist.
A hint
An emitted electron cannot have negative kinetic energy.
See the worked solution

Photon energy is insufficient to free the electron: 2,0 − 2,4 = −0,4 eV is not emission kinetic energy. The result indicates no emission in this one-photon-per-electron regime. Increasing only intensity while keeping individual energy unchanged does not overcome that threshold within the model.

Exercise 3

An atom emits a 3 eV photon in a transition. The final state has energy −5 eV. What was the initial energy?
A hint
In emission, photon E = initial E − final E.
See the worked solution

3 = initial E − (−5), so initial E = −2 eV. Initial energy was greater than final energy. Both levels being negative does not prevent their emitted difference from being positive.

Exercise 4

After two half-lives, what expected fraction remains? Can we predict which individual nucleus will transform next?
A hint
The exponential law predicts the population and probabilities, not an order of events.
See the worked solution

On average, (1/2)² = 1/4 remains. This law alone does not predict the identity of the next nucleus. For a small population, substantial relative deviations around the expected fraction may be observed.

Exercise 5

In alpha decay, a nucleus with Z = 84 and A = 212 produces which Z and A values in the residual nucleus?
A hint
The alpha particle carries away two protons and two neutrons.
See the worked solution

Residual Z = 84 − 2 = 82 and residual A = 212 − 4 = 208. The residual neutron number is A − Z = 126. This numerical identification respects charge and nucleon balance in the transformation.

Exercise 6

In a reaction, the total rest mass before the reaction exceeds the total rest mass of the products by 2.0 × 10⁻⁹ kg. How much energy is released? Use c = 3.0 × 10⁸ m/s.
A hint
The square also applies to the power of ten in c.
See the worked solution

Q = Δmc² = 2,0 × 10⁻⁹ × 9,0 × 10¹⁶ = 1,8 × 10⁸ J. Energy must be accounted for in products and radiation; the relationship is not an instruction to convert any object entirely into useful energy.

Exercise 7

Can a massive particle reach c if it receives a sufficiently large finite amount of energy?
A hint
Examine the denominator of γ as v approaches c.
See the worked solution

Not in the relativistic model. The term 1 − v²/c² tends to zero from positive values and γ grows without limit. The energy required also grows without limit. This statement concerns acceleration of a massive particle; photons have zero rest mass and their own energy/momentum relationship.

Mistakes worth avoiding

A new theory makes everything that came before useless.
Classical models remain useful in regimes where they are good approximations; extended theories explain their limits.
More intensity gives more energy to each photon.
Individual energy depends on frequency. Intensity can change the average number of incident photons.
An orbital is an imprecisely drawn planetary orbit.
A quantum electronic state describes energies and probability distributions; it is not simply an unknown classical trajectory.
After two half-lives the original sample disappears.
The expected remaining value is one quarter. The law describes a radionuclide's population and allows statistical fluctuations.
Activity directly measures all radiation energy or every radiation effect.
Becquerels measure transformations per second. Energy per event, products, material and geometry are additional information.
E = mc² means kinetic energy is always mc².
mc² is rest energy. For a free massive particle, total E = γmc² and K = (γ − 1)mc².

The essentials to take with you

  • Modern physics extends models when new scales and experiments require it.
  • A photon has E = hf; beam energy also depends on how many photons it contains.
  • The photoelectric effect requires sufficient energy to overcome the work function in the simple model.
  • Bound energy levels and transitions help interpret atomic spectra.
  • Matter waves and probabilities have experimental evidence; they are not merely poetic images.
  • Nuclear transformations differ from electronic transitions and ionization.
  • Half-life and activity describe populations and rates, with clear units and conditions.
  • Energy and momentum are also conserved in the appropriate quantum and relativistic descriptions.

If you still cannot explain one of these ideas, return to the corresponding section and work through an example again. Recognising the difficulty is part of studying.

References for further study

Text and examples prepared for this learning path. These sources help check the principles and continue studying:

The grade organization is a suggested study sequence. Grade 13 provides general extension and preparation for higher education. Pedagogical review by teachers remains a stage of the project.