Chen C.J. Physics of Solar Energy

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62  Nature of Solar Radiation

Figure 2.7 Albert Einstein and Robert Millikan. Both Einstein and Millikan won a Nobel Prize

for their contributions to the photoelectric eﬀect. Photograph taken in 1930 when Robert Millikan

invited Albert Einstein to a conference in California. Original photograph courtesy of Smithsonian

Museum, slightly cleaned up by the author.

present as resting upon any sort of a satisfactory theoretical foundation [61].” In 1950,

at age 82, in his autobiography [62], Millikan reversed his position and admitted that

his experiments

proved simply and irrefutably, I thought, that the emitted electron that

escapes with the energy hν gets that energy by the direct transfer of hν

units of energy from the light to the electron, and hence scarcely permits of

any other interpretation than that which Einstein had originally suggested,

namely that of the semi-corpuscular or photon theory of light itself.

2.4.3 Wave–Particle Duality

The earlier objections to Einstein’s theory of photons was related to an even more

profound problem: the wave–particle duality of all particles. At the beginning of the

twentieth century, electrons were described by classical mechanics as being similar to

billiard balls. Einstein’s theory seemed to imply that photons are also like billiard balls

and that the photoelectric eﬀect is a collision of the billiard balls with electrons. Such a

picture is not only hard to conceive but also in direct conﬂict with the well-established

interference phenomenon of light.

The paradox was resolved after Louis de Broglie extended Einstein’s postulate that

light can be both wave and particle to all particles, including the electron. According

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2.5 Einstein’s Derivation of Blackbody Formula  63

to de Broglie, a particle with a momentum p should also be a plane wave with a

wavevector k such that

p =  k, (2.114)

where  = h/2π, Planck’s constant divided by 2π, is often called Dirac’s constant.

According to the theory of de Broglie, a better picture of the photoelectric eﬀect is

that light radiation as a plane wave interacts with the electron in the electrode, which

is also a plane wave, but the energy transferred to the electron must be quantized to

satisfy the Einstein equation

 = hν ≡ ω, (2.115)

where ω =2πν is the circular frequency of the light wave. This is an essential concept

regarding the understanding of solar cells and solar photochemistry, which we will

discuss in the corresponding chapters.

2.5 Einstein’s Derivation of Blackbody Formula

Based on the concept of photons and the interaction of photons with matter, Einstein

made a very simple derivation of the blackbody radiation formula. The key of his

derivation is the introduction of stimulated emission, which gave birth to the laser,

an acronym for light ampliﬁcation by stimulated emission of radiation, and provides a

better understanding of the interaction between solar radiation and atomic systems.

Einstein studied a simple two-state atomic system; see Fig. 2.8. The radiation

ﬁeld is represented by an energy density ρ(ν), where ν is the frequency. The atomic

system has two states with an energy diﬀerence hν. According to Maxwell–Boltzmann

statistics, the ratio of the populations of the two states is

= e

−hν/k

. (2.116)

Einstein assumed three transition coeﬃcients: the absorption coeﬃcient B

,the

spontaneous emission coeﬃcient A, and the stimulated emission coeﬃcient B

.The

rate equations are

= B

ρ(ν) − B

ρ(ν) − AN

, (2.117)

= −B

ρ(ν)+B

ρ(ν)+AN

. (2.118)

At equilibrium, both dN

/dt and dN

/dt should vanish. Therefore,

ρ(ν)

A + B

ρ(ν)

= e

−hν/k

. (2.119)

The coeﬃcients should not depend on temperature. At high temperature, the power

density should be high, and the right-hand side of Eq. 2.119 should approach unity.

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64  Nature of Solar Radiation

Figure 2.8 Einstein’s derivation of blackbody

radiation formula. The radiation ﬁeld ρ(ν) inter-

acts with a two-level atomic system. Three inter-

action modes are assumed: absorption, to lift the

atomic system from state 1 to state 2; spontaneous

emission and stimulated emission, the atomic system

decays from state 2 to state 1, giving out energy to

the radiation ﬁeld.

Therefore, one must have

= B

= B. (2.120)

The absorption coeﬃcient B

equals the stimulated emission coeﬃcient B

,whichcan

be represented by a single coeﬃcient B. Under any temperature, the power density

distribution of radiation is then

ρ(ν)=

hν/k

− 1

. (2.121)

For radiations of low photon energy, Eq. 2.121 reduces to

ρ(ν) →

hν

. (2.122)

It should be identical to the Rayleigh–Jeans formula. Comparing with Eq. 2.90, we

ﬁnd the ratio of coeﬃcients A and B,

8πhν

. (2.123)

Finally, Planck’s formula is recovered,

ρ(ν)=

8πhν

hν/k

− 1

. (2.124)

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Problems

2.1. Show that the capacitance C of a parallel-plate capacitor with vacuum as the

dielectric is

C =

[F], (2.125)

where A is the area and d is the distance between the electrodes.

2.2. Show that the capacitance C of a parallel-plate capacitor with a medium of relative

dielectric constant ε

C =

[F]. (2.126)

Calculate the capacitance of a capacitor with A =1m

and s = 1 mm for glass and

silicon.

2.3. Show that the inductance L of an inductor made of a long solenoid of N loops

with cross-sectional area A and length l is

L =

[H]. (2.127)

2.4. Show that the speed of light v in a medium of relative dielectric constant ε

v =

√

. (2.128)

Calculate the speed of light v in glass and silicon (the relative dielectric constants ε

for glass and silicon are 2.25 and 11.7, respectively).

2.5. The refractive index of window glass is n =1.50. How much light power is lost

when going through a sheet of glass at normal incidence? (Hint: there are two glass–air

interfaces.)

2.6. The radius of the Sun is R =6.96 ×10

m, and the distance between the Sun and

Earth is D =1.5 × 10

m. The solar constant is 1366 W/m

. Estimate the surface

temperature of the Sun. (Hint: use the Stefan–Boltzmann law.)

2.7. What is the magnitude of the electric ﬁeld intensity of the sunlight just outside

the atmosphere of Earth?

2.8. What is the electric ﬁeld intensity of the electron in a hydrogen atom at the

distance of one Bohr radius from the proton?

2.9. Derive the blackbody radiation spectral density per unit wavelength in unit of

micrometers.

2.10. Using the blackbody radiation formula per unit wavelength, derive the Wien

displacement law in micrometers.

Problems  65

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66  Nature of Solar Radiation

Figure 2.9 Wavelengths of visible lights.

2.11. The wavelengths of visible light with diﬀerent colors in nanometers are shown

in Fig. 2.9. Compute the frequencies and energy values of the photons, in both joules

and electron volts.

2.12. What is the solar constant of Venus? Assume that the Sun is a blackbody emitter

at 5800 K and the mean Venus-Sun distance is 1.08 × 10

2.13. To compute the blackbody irradiation for photon energy from 

to inﬁnity, an

easy-to-use formula can be obtained by introducing x

= 

T and expanding the

denominator of Eq. 2.99 into

U(T,

2π(k

T )



∞

−x

1 − e

−x

2π(k

T )

∞



n=1



∞

−nx

dx.

(2.129)

Prove that

U(T,

2π(k

T )

∞



n=1

−nx





, (2.130)

with



. (2.131)

2.14. Assuming that the Sun is a blackbody emitter at 5800 K, what fraction of solar

radiation is green (wavelength between 495 and 570 nm)?

2.15. Assuming that the Sun is a blackbody emitter at 5800 K, what fraction of solar

radiation has photon energy greater than 1.1 eV?

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Chapter 3

Origin of Solar Energy

Since the late-nineteenth century, measurement of solar radiation power density has

been diligently pursued. After the advancement of artiﬁcial satellites, solar radiation

data outside the atmosphere became available. From the data accumulated to date,

variation in the solar radiation power density at the average position of Earth outside

the atmosphere has not exceeded 0.1% over a century. Although not a physical constant,

this quantity, S, is often called the solar constant,

S = 1366 ± 3W/m

. (3.1)

The total radiation power of the Sun, L



,orthesolar luminosity, can be evaluated

using the solar constant and the average distance between the Sun and Earth, A



,the

astronomical unit of length,



=1.5 × 10

m, (3.2)

which gives (see Fig. 3.1)



=4πA



S =3.84 × 10

W. (3.3)

Figure 3.1 Luminosity of the Sun. The average radiation power density of sunlight outside the

atmosphere of Earth, the solar constant,isS = 1366 W/m

. The average distance between the Sun

and Earth, the astronomical constant,isA



=1.5 × 10

m. The luminosity of the Sun, or the total

power of solar radiation, is then L



=3.84 × 10

Physics of Solar Energy C. Julian Chen

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68  Origin of Solar Energy

The total power of solar radiation is about twenty trillion times the energy consumption

of the entire world, 1.6 ×10

W. For the purpose of utilizing solar poinof solar radiation

can one expect?

3.1 Basic Parameters of the Sun

For centuries, the Sun has been the subject of intense study by astronomers. The basic

parameters and the basic structure of the Sun are well understood [7, 63, 68, 79]. Here

is a brief summary.

3.1.1 Distance

The distance between the Sun and Earth has been measured using triangulation and

radar echoes. Because the orbit of Earth around the Sun is an ellipse, the distance is not

constant. Around January 3, at perihelion, the distance is at a minimum, 1.471 ×10

m. Around July 3, at aphelion, the distance is at a maximum, 1.521 × 10

m. The

average distance, A



=1.5 × 10

m (see Eq. 3.2) is treated as a basic parameter in

astronomy.

Besides the distance in meters, the average time for light to travel between the Sun

and Earth, the light time for the astronomical unit of distance τ



, has been determined

with high accuracy. For all practical applications in solar energy, the approximate

value, accurate to 0.2%, is worth memorizing:



= 500 s. (3.4)

3.1.2 Mass

The mass of the Sun is measured using Kepler’s law and the orbital parameters of the

planets. Again, for all practical applications, an approximate value, accurate to 1%, is

worth memorizing:



=2× 10

kg. (3.5)

It is 333,000 times the mass of Earth. Due to radiation and solar wind, the Sun loses

kg of its mass per year, which is negligible. Practically, the mass of the Sun is a

constant over its lifetime.

3.1.3 Radius

The radius of the Sun is measured by the angular diameter of the visible disc. Because

the distance between the Sun and Earth is not constant, the angular diameter of the

Sun is not constant. It ranges from 31.6



to 32.7



.Theaveragevalueis32



, or 0.533

◦

Using the average distance A



, the radius of the Sun is then

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3.1 Basic Parameters of the Sun  69



1.5 × 10

× 0.533

2 × 57.3

∼

6.96 × 10

m. (3.6)

Calculated from the radius, the volume of the Sun is 1.412×10

, and its average

density is 1.408 g/cm

3.1.4 Emission Power

The emission power of the surface of the Sun, in watts per square meter, can be

calculated from Eqs. 3.3 and 3.6:



4πr



3.84 × 10

4π (6.96 × 10

)

∼

63.1MW/m

. (3.7)

3.1.5 Surface Temperature

Considering the Sun as a blackbody, the temperature can be calculated from the Stefan–

Boltzmann law. Using Eqs. 3.7, 2.99, and 2.101,









1/4



63.1 × 10

5.67 × 10

−8



1/4

∼

5800 K. (3.8)

In the literature, the surface temperature of the Sun varies from one source to another,

from 5600 to 6000 K. The diﬀerence is insigniﬁcant. Throughout this book, we will

use 5800 K as the nominal surface temperature of the Sun. Not only is this easy to

remember, but it is exactly equal to 0.5 eV, which is convenient for the treatment of

solar cells.

Table 3.1: Chemical Composition of the Sun

Element Z Molecular Abundance (% of Abundance

Weight Number of Atoms) (% of Mass)

Hydrogen 1 1.008 91.2 71.0

Helium 2 4.003 8.7 27.1

Oxygen 8 16.000 0.078 0.97

Carbon 6 12.011 0.043 0.40

Nitrogen 7 14.007 0.0088 0.096

Silicon 14 28.086 0.0045 0.099

Magnesium 12 24.312 0.0038 0.076

Neon 10 20.183 0.0035 0.058

Iron 26 55.847 0.0030 0.14

—emphSource: Ref. [79].

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70  Origin of Solar Energy

3.1.6 Composition

The spectrum of solar radiation does not precisely match the spectrum of blackbody

radiation. The ﬁne structure of the solar spectrum provides evidence of its chemical

composition (Table 3.1). Actually, the second most abundant element in the Sun,

helium, which was discovered from the spectrum of the solar radiation, comes from the

Greek word helios.

3.2 Kelvin–Helmholtz Time Scale

For centuries, the origin of sunlight has been a fundamental inquiry of all cultures. In

the middle of the nineteenth century, the ﬁeld of thermodynamics matured. The ﬁrst

law of thermodynamics states that energy can only change its form and can neither be

created nor be destroyed. The immense solar radiation is apparently draining enormous

amounts of energy from the Sun. In the 1850s, Sir William Thomson (Fig. 3.2), one

of the founders of thermodynamics, made a thorough study on the origin of the Sun’s

energy based on the knowledge of physics at that time [81]. He argued that chemical

energy could not be the answer, because even if the Sun is made of coal and burns

completely, it can only supply a few thousand years of sunlight. The only explanation

based on the knowledge of that time was the gravitational contraction of the Sun, also

independently proposed by German physicist Hermann von Helmholtz; see Fig. 3.3(a).

When two meteorites with masses m

and m

approach from inﬁnity to a distance r,

Figure 3.2 Sir William Thomson.

Irish-born Scottish physicist (1824-

1907), aka Lord Kelvin, one of the

founders of thermodynamics. He at-

tributed the origin of the Sun’s energy

to gravitational interaction, and as-

serted that the Sun could not shine for

more than 30 million years. He used

his theory to refuse Sir Charles Lyell’s

geology, and especially Charles Dar-

win’s evolutionary biology; both re-

quire that the lifetime of the Sun must

be at least one billion years. The dis-

crepancy was resolved by Albert Ein-

stein and Hans Bethe in the 20th cen-

tury. Artwork by Hubert Herkomer,

courtesy of Smithsonian Museum.

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3.2 Kelvin–Helmholtz Time Scale  71

Figure 3.3 The Kelvin-Helmholtz model. A model of the origin of solar energy based on

gravitational contraction. (a) When two meteorites approach each other from faraway, the gravitational

potential energy is reduced and the kinetic energy is increased. (b) If the Sun is formed by coalition of

a large number of meteorites, the gravitational energy can be converted into kinetic energy, and then

radiated as heat.

the potential energy is decreased by

U = −

, (3.9)

where G =6.67 × 10

−11

N · m

/kg

is the gravitational constant. Because of gravita-

tional attraction, the system gains a kinetic energy

. (3.10)

Assuming that the Sun is formed by the coalescence of large number of meteorites

into a sphere of radius r



, as shown in Fig. 3.3(b), kinetic energy can be gained by

gravitational attraction. The value of the energy gain depends on the distribution of

the mass. By analogy to Eq. 3.10, we can make a good order-of-magnitude estimate

of the energy gain,

∼



. (3.11)

Obviously it is an overestimate. For example, by calculating the energy gain with a

uniform mass distribution, a numerical factor of 0.6 appears. However, Eq. 3.11 gives

an upper limit. Based on the Kelvin–Helmholtz assumption, the lifetime of the Sun

can be estimated by dividing that energy gain by the rate of energy loss, or the solar

luminosity L



∼



6.67 × 10

−11



2 × 10



6.96 × 10

× 3.84 ×10

≈ 10

s. (3.12)

The quantity τ

, about 30 million years, is called the Kelvin–Helmholtz time scale.

Based on this calculation, Lord Kelvin asserted that the Sun’s life span must not exceed

30 million years [81].