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Radiation from real bodies IB DP Physics Study Notes

Radiation from real bodies IB DP Physics Study Notes - 2025 Syllabus

Radiation from real bodies IB DP Physics Study Notes

Radiation from real bodies 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 conservation of energy

  • emissivity as the ratio of the power radiated per unit area by a surface compared to that of an ideal black surface at the same temperature as given by emissivity $=\frac{\text { power radiated per unit area }}{\sigma T^4}$

  • albedo as a measure of the average energy reflected off a macroscopic system as given by

    $\text { albedo }=\frac{\text { total scattered power }}{\text { total incident power }}$

  • that Earth’s albedo varies daily and is dependent on cloud formations and latitude

  • the solar constant $S$

  • that the incoming radiative power is dependent on the projected surface of a planet along the direction of the path of the rays, resulting in a mean value of the incoming intensity being $\frac{S}{4}$

Standard level and higher level: 6 hours
Additional higher level: There is no additional higher level content

IB DP Physics 2025 -Study Notes -All Topics

emissivity

The emissivity e of a body is a number between 0 and 1 that quantifies the emission and absorption properties of that body as compared to a blackbody of equal size.

  • A black-body is a perfect emitter / absorber of radiation and has an emissivity of e = 1.
  • A body that can’t emit / absorb radiation at all has an emissivity of e = 0.
  • The emissivity of a body is the ratio of the power emitted by the body, to the power emitted by a black-body of the same size.
  • $e=\frac{P_{BODY}}{\sigma T^4 }$
  • A black-body is a perfect emitter/absorber (e = 1).
  • A small sphere covered with lamp black makes a pretty good black-body and has an emissivity somewhat close to 1.
  • An even better black- body is the metal-cavity black-body:
  • If we know e, the Stefan-Boltzmann law can be amended:

Albedo

  • When light strikes an object, some of it is absorbed, and some of it is scattered.
  • If light strikes a mirror, nearly all of it will be scattered:
  • If light strikes a surface covered with lamp black, nearly all of it will be absorbed:
  •  We define albedo in terms of scattered and incident power:
FYI
  • The mirror has an albedo of almost 1.
  • The black-body has an albedo of 0.

Landforms, vegetation, weather, and seasons affect a planet’s albedo.

  • The table shows  that different  terrains have different albedos.
  • Ocean water scatters little light (7%).
  • Snow and ice scatter a lot of light (62% to 66%).
  • Desert (36%).

Satellites are used to map out monthly albedos.

  • From the previous table and satellite map it is clear that calculating the overall albedo of a planet is quite complex.
  • Clouds also contribute to the albedo.
  • Overall, Earth’s mean yearly albedo is about 0.3 (or 30%).
  • The actual albedo depends on season, latitude, cloud cover, and snow cover.
  • It varies daily!

EXAMPLE: Assuming an albedo of 0.30 , find, for Earth:

(a) the power of the sunlight received.

(b) the predicted temperature due to the sunlight reaching it.

▶️Answer/Explanation

SOLUTION:

(a) Use $I=1380 \mathrm{Wm}^{-2}$ and $I=\frac{P}{A}$.
-The radius of Earth is $r=6.37 \times 10^6 \mathrm{~m}$ so its cross-sectional area is

$$
A=\pi r^2=\pi\left(6.37 \times 10^6\right)^2=1.27 \times 10^{14} \mathrm{~m}^2
$$

-An albedo of 0.30 means that $70 \%$ of the sunlight is absorbed (because $30 \%$ is scattered). Thus

$$
P=(0.70) I A=(0.70)(1380)\left(1.27 \times 10^{14}\right)=1.23 \times 10^{17} \mathrm{~W}
$$

Thus Earth intercepts energy from the sun at a rate of $1.23 \times 10^{17} \mathrm{~W}$.

(b)

 $P=1.23 \times 10^{17} \mathrm{~W}$.
But this power is distributed over the whole rotating planet, which has an area $A_{\text {sphere }}=4 \pi r^2$.
From Stefan-Boltzmann we have

$$
\begin{aligned}
& P=\sigma A T^4 \\
& 1.23 \times 10^{17}=\left(5.67 \times 10^{-8}\right) 4 \pi\left(6.37 \times 10^6\right)^2 T^4 \\
& T=255 \mathrm{~K}\left(-18^{\circ} \mathrm{C}\right)
\end{aligned}
$$

The solar constant

The sun radiates energy at a rate of $3.90 \times 10^{26} \mathrm{~W}$. What is the rate at which energy from the sun reaches Earth if our orbital radius is $1.5 \times 10^{11} \mathrm{~m}$ ?

$A_{\text {SPHERE }}=4 \pi r^2$.

Recall that intensity is the rate at which energy is being gained per unit area.

Then $I=\frac{r}{4 \pi r^2}$

$$
=\frac{3.90 \times 10^{26}}{4 \pi\left(1.5 \times 10^{11}\right)^2}=1380 \mathrm{~W} \mathrm{~m}^{-2} .
$$

This value is called the solar constant.

$$
P_{\text {sun }}=1380 \mathrm{~W} \mathrm{~m}^{-2} \quad \text { the solar constant }
$$

PRACTICE: 

Explain why the solar intensity is different for different latitudes. 

▶️Answer/Explanation

SOLUTION:

∙The following diagram shows how the same intensity is spread out over more area the higher the latitude.

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