Q1 Thermal Radiation, Nuclear Fusion, and HR Diagrams

Solar Radiation, Greenhouse Effect, Fusion, and Stellar Evolution

IB Physics HL • Paper 2 • Long-Answer Question • Astrophysics, Thermal Radiation, Nuclear Fusion, and HR Diagrams

Lesson Overview

In this Paper 2 long-answer question, we study how solar radiation reaches Earth, how Earth’s surface–atmosphere system maintains energy balance, and how greenhouse gases affect the re-radiation of energy. The question then develops into nuclear fusion in the Sun, mass defect and energy release in the proton–proton chain, and ends with stellar classification and fusion processes in a massive star such as Antares.

This question is a strong example of how IB Physics combines thermal physics, astrophysics, and nuclear physics in one multi-step problem. To score well, you must show the physical principle first, then the calculation, then a short concluding statement that answers the question directly.

⭐ Key Concepts

The solar constant is the intensity of solar radiation received at Earth.
Earth reflects part of the incoming solar radiation. The fraction reflected is called the albedo.
A black-body or grey-body surface emits thermal radiation according to

$I=\epsilon \sigma T^4$

Greenhouse gases absorb infrared radiation when photon energies match molecular energy-level differences.
Absorbed radiation is re-emitted in random directions, including back towards Earth’s surface.
The total power radiated by a star can be related to intensity at distance $r$ using

$P=4\pi r^2 I$

In nuclear fusion, the energy released comes from the mass defect:

$E=\Delta mc^2$

On the HR diagram, a star’s position shows its luminosity and temperature, helping identify its type and evolutionary stage.

📘 Clear IB-Style Explanation

This question is built around the idea of energy flow. First, solar energy arrives at Earth. Some is reflected because of albedo, some is absorbed by the surface, and some is emitted again as thermal radiation. The atmosphere then changes this balance because greenhouse gases absorb and re-radiate infrared radiation.

Later parts of the question shift to the Sun as the source of that energy. In IB Physics, it is important to connect:

solar radiation to energy balance on Earth,
mass defect to energy released in fusion,
fusion in stars to stellar structure and evolution.

Strong answers to these questions combine precise definitions, correct equations, and short scientific explanations using clear cause-and-effect language.

📌 Worked Example 1 — Defining the Solar Constant

(a) State what is meant by the solar constant.

Solution

The solar constant is the intensity of solar radiation received at Earth.

It may also be described as the power per unit area received from the Sun at Earth’s distance.

📊 Marks: A1

📌 Worked Example 2 — Greenhouse Gases and Energy Absorbed by Earth’s Surface

The following data are given:

$\text{Average albedo of Earth}=0.30$

$\text{Average global temperature of the surface}=288\text{ K}$

$\text{Average Earth-Sun distance}=1.5\times10^{11}\text{ m}$

(b)(i) Outline the physical mechanism by which some of the radiation emitted by the surface is absorbed by greenhouse gases in the atmosphere and re-radiated towards the surface.

(b)(ii) Show that the average global intensity of radiation absorbed by the surface is about $240\text{ W m}^{-2}$ .

Solution

(b)(i) Mechanism of greenhouse gas absorption and re-radiation

Greenhouse gas molecules absorb infrared radiation when the photon energy matches the difference between molecular energy levels.

This may also be described as resonance absorption by the molecule.

After absorbing the radiation, the greenhouse gas molecules re-emit radiation in random directions.

Some of this re-radiated energy travels back towards Earth’s surface.

📊 Marks: A1 A1

(b)(ii) Average global intensity absorbed by the surface

The solar constant is approximately

$S=1360\text{ W m}^{-2}$

Because Earth is a sphere but intercepts sunlight over a circular cross-section, the average incoming intensity over the whole Earth is

$\frac{S}{4}$

$\frac{1360}{4}=340\text{ W m}^{-2}$

Earth reflects $30\%$ of the incoming radiation, so the fraction absorbed is

$1-0.30=0.70$

Hence the average global intensity absorbed is

$0.70\times340$

$=238\text{ W m}^{-2}$

So the average absorbed intensity is about

$2.4\times10^2\text{ W m}^{-2}$

📊 Marks: M1 A1

📌 Worked Example 3 — Radiation Re-radiated by the Atmosphere

(b)(iii) Determine the average intensity re-radiated by the atmosphere towards the surface. Assume that the emissivity of the surface is $0.90$ .

Solution

The intensity emitted by the Earth’s surface is given by the Stefan–Boltzmann law:

$I=\epsilon \sigma T^4$

Substitute

$\epsilon=0.90,\qquad \sigma=5.67\times10^{-8}\text{ W m}^{-2}\text{K}^{-4},\qquad T=288\text{ K}$

$I=(0.90)(5.67\times10^{-8})(288)^4$

$I\approx351\text{ W m}^{-2}$

The outgoing intensity from the Earth–atmosphere system is about

$238\text{ W m}^{-2}$

So the intensity re-radiated by the atmosphere towards the surface is

$351-238$

$=113\text{ W m}^{-2}$

Therefore, the average intensity re-radiated by the atmosphere towards the surface is

$113\text{ W m}^{-2}$

📊 Marks: M1 M1 A1

📌 Worked Example 4 — Total Power Radiated by the Sun

(c) Show, with reference to the solar constant, that the total power radiated by the Sun is about $4\times10^{26}\text{ W}$ .

Solution

At Earth’s orbit, the Sun’s radiation is spread uniformly over the surface area of a sphere of radius

$r=1.5\times10^{11}\text{ m}$

So the total power radiated by the Sun is

$P=4\pi r^2 S$

Substitute the values:

$P=4\pi(1.5\times10^{11})^2(1360)$

$P=4\pi(2.25\times10^{22})(1360)$

$P\approx3.85\times10^{26}\text{ W}$

So the total power radiated by the Sun is about

$4\times10^{26}\text{ W}$

📊 Marks: M1 A1

📌 Worked Example 5 — Energy Released in the Proton–Proton Chain

The primary energy source of the Sun is the proton–proton chain of fusion reactions. Four protons and two electrons produce a helium nucleus together with neutrinos and gamma photons.

(d)(i) The mass of the helium nucleus is $4.001506\text{ u}$ . Calculate, in MeV, the energy released in the reaction.

Solution

The mass defect is

$\Delta m=4(1.007276)+2(0.000549)-4.001506$

$\Delta m=4.029104+0.001098-4.001506$

$\Delta m=0.028696\text{ u}$

Using

$1\text{ u}=931.5\text{ MeV }c^{-2}$

the energy released is

$E=\Delta mc^2$

$E=(0.028696)(931.5)$

$E=26.7\text{ MeV}$

Therefore, the energy released is

$26.7\text{ MeV}$

📊 Marks: M1 A1

📌 Worked Example 6 — Fusion and the Stability of the Sun

(d)(ii) Outline the role of fusion reactions in maintaining a stable radius of the Sun.

Solution

Fusion reactions in the Sun release energy in the form of photons.

These photons contribute to outward radiation pressure and thermal pressure inside the Sun.

This outward pressure balances the inward gravitational force that would otherwise compress the Sun.

As a result, the Sun maintains an approximately stable radius.

📊 Marks: A1 A1

📌 Worked Example 7 — Evidence for Helium in the Sun

(d)(iii) Outline how the presence of helium in the Sun can be confirmed empirically.

Solution

The presence of helium can be confirmed by studying the Sun’s spectrum.

If the absorption or emission lines in the solar spectrum match the known spectral lines of helium, then helium is present in the Sun.

📊 Marks: A1 A1

📌 Worked Example 8 — Antares on the HR Diagram

The positions of the Sun and the star Antares are shown in the Hertzsprung–Russell (HR) diagram.

(e)(i) State the star type of Antares.

(e)(ii) Discuss how nuclear fusion processes in Antares are different from those in the Sun.

Solution

(e)(i) Star type of Antares

Antares is a

$\text{red supergiant}$

📊 Marks: A1

(e)(ii) Fusion in Antares compared with the Sun

The Sun mainly fuses hydrogen into helium in its core.

Antares, being a much more massive and more evolved star, can fuse elements heavier than hydrogen. In later stages, fusion can continue up to iron.

Fusion in Antares occurs at a much greater rate because its core is hotter and denser.

Also, Antares can have different fusion regions or shells, where different elements are being fused at the same time.

Therefore, nuclear fusion in Antares is both more complex and more advanced than in the Sun.

📊 Marks: A1 A1 A1

⚠ Common Mistakes

Defining the solar constant as total power rather than intensity at Earth.
Forgetting to divide the solar constant by 4 when finding the global average intensity over Earth.
Using albedo incorrectly by subtracting the wrong fraction.
Confusing radiation emitted by Earth’s surface with radiation leaving the whole Earth–atmosphere system.
Using the Stefan–Boltzmann law without including the emissivity factor.
Omitting the two electrons in the proton–proton chain mass-defect calculation.
Saying helium is “seen in the Sun” without referring to spectral lines.
Calling Antares a red giant instead of a red supergiant.

📘 IB Exam Tips

For definitions, give the shortest precise scientific statement possible.
In global radiation-balance questions, always think carefully about geometry: cross-sectional area versus surface area of a sphere.
When using

$I=\epsilon \sigma T^4$

make sure temperature is in kelvin.
For fusion questions, show the mass defect clearly before converting to energy.
For spectrum questions, always mention matching observed lines to known element lines.
For stellar evolution questions, link the star’s mass and stage of evolution to the type of fusion taking place.

🧪 Challenge Problem

A planet receives solar radiation of intensity $1200\text{ W m}^{-2}$ at the top of its atmosphere. The albedo of the planet is $0.25$ .

(a) Calculate the average global intensity of solar radiation absorbed by the planet.

(b) The surface temperature is $270\text{ K}$ and the emissivity of the surface is $0.80$ . Calculate the intensity emitted by the surface.

(c) If the outgoing intensity from the planet–atmosphere system is $225\text{ W m}^{-2}$ , determine the average intensity re-radiated by the atmosphere towards the surface.

(d) Explain why greenhouse gases are able to absorb infrared radiation.

✅ Self-Practice

A star radiates uniformly in all directions. The intensity of radiation measured at a distance of $2.0\times10^{11}\text{ m}$ is $950\text{ W m}^{-2}$ .

(a) Calculate the total power radiated by the star.

(b) In a fusion reaction, the total mass before the reaction is $4.0312\text{ u}$ and the total mass after the reaction is $4.0026\text{ u}$ . Calculate the energy released in MeV.

(c) State how the presence of a particular element in a star can be confirmed using its spectrum.

(d) A red supergiant has reached a later stage of evolution than the Sun. State one way in which fusion in this star differs from fusion in the Sun.

Show Solutions

Challenge Problem

(a)

Average incoming intensity over the whole planet:

$\frac{1200}{4}=300\text{ W m}^{-2}$

Fraction absorbed:

$1-0.25=0.75$

So the absorbed intensity is

$0.75\times300=225\text{ W m}^{-2}$

(b)

$I=\epsilon \sigma T^4$

$I=(0.80)(5.67\times10^{-8})(270)^4$

$I\approx241\text{ W m}^{-2}$

(c)

$241-225=16\text{ W m}^{-2}$

So the atmosphere re-radiates

$16\text{ W m}^{-2}$

towards the surface.

(d)

Greenhouse gases absorb infrared radiation when the photon energy matches the energy difference between molecular energy levels. The molecules then re-emit radiation in random directions.

Self-Practice

(a)

$P=4\pi r^2 I$

$P=4\pi(2.0\times10^{11})^2(950)$

$P\approx4.8\times10^{26}\text{ W}$

(b)

Mass defect:

$\Delta m=4.0312-4.0026=0.0286\text{ u}$

Energy released:

$E=\Delta mc^2=(0.0286)(931.5)$

$E\approx26.6\text{ MeV}$

(c)

The element is identified by comparing the absorption or emission lines in the star’s spectrum with the known spectral lines of that element.

(d)

A red supergiant can fuse heavier elements than hydrogen, whereas the Sun mainly fuses hydrogen into helium.

Solar Radiation, Greenhouse Effect, Fusion, and Stellar Evolution

Lesson Overview

⭐ Key Concepts

📘 Clear IB-Style Explanation

📌 Worked Example 1 — Defining the Solar Constant

Solution

📌 Worked Example 2 — Greenhouse Gases and Energy Absorbed by Earth’s Surface

Solution

📌 Worked Example 3 — Radiation Re-radiated by the Atmosphere

Solution

📌 Worked Example 4 — Total Power Radiated by the Sun

Solution

📌 Worked Example 5 — Energy Released in the Proton–Proton Chain

Solution

📌 Worked Example 6 — Fusion and the Stability of the Sun

Solution

📌 Worked Example 7 — Evidence for Helium in the Sun

Solution

📌 Worked Example 8 — Antares on the HR Diagram

Solution

⚠ Common Mistakes

📘 IB Exam Tips

🧪 Challenge Problem

✅ Self-Practice

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Q1 Thermal Radiation, Nuclear Fusion, and HR Diagrams

Solar Radiation, Greenhouse Effect, Fusion, and Stellar Evolution

Lesson Overview

⭐ Key Concepts

📘 Clear IB-Style Explanation

📌 Worked Example 1 — Defining the Solar Constant

Solution

📌 Worked Example 2 — Greenhouse Gases and Energy Absorbed by Earth’s Surface

Solution

📌 Worked Example 3 — Radiation Re-radiated by the Atmosphere

Solution

📌 Worked Example 4 — Total Power Radiated by the Sun

Solution

📌 Worked Example 5 — Energy Released in the Proton–Proton Chain

Solution

📌 Worked Example 6 — Fusion and the Stability of the Sun

Solution

📌 Worked Example 7 — Evidence for Helium in the Sun

Solution

📌 Worked Example 8 — Antares on the HR Diagram

Solution

⚠ Common Mistakes

📘 IB Exam Tips

🧪 Challenge Problem

✅ Self-Practice

Related

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