1.

5 Electromagnetic Radiations and Atomic Spectra

1.5 .1 Electromagnetic Radiations

James Maxwell (1873) was the first to give a comprehensive explanation about the interaction
between the charged bodies and the behavior of electrical and magnetic fields on macroscopic
level. He suggested that when electrically charged particle moves under acceleration, alternating
electrical and magnetic fields are produced and transmitted. These fields acting together
produced radiant energy that transmitted in the forms of waves called electromagnetic waves or
electromagnetic radiation. .

According to Maxwell’s theory, visible light consists of electromagnetic waves. An

electromagnetic wave has an electric filed component and a magnetic field component. Waves
are characterized by their length and height and by the number of waves that pass through a
certain point in one second .

Wavelength (lambda , λ) is the distance between identical points on successive waves. It is

expressed in meters (m) and often, for very short wavelengths, in nanometers (nm), picometers
(pm), or angstrom (Å .The frequency (v) is the number of waves that pass through a particular
point in 1 second, and is expressed in units of l/second (1/s; also called hertz, Hz).
Amplitude is the vertical distance from the midline of a wave to the peak or trough.

Figure 1.7, Page 16: Frequency of waves(draw it)

Another important property of waves is their speed, which depends on the type of wave and the
nature of the medium through which the wave is traveling (for example, air, water, or a vacuum).
The speed (u) of a wave is the product of its wavelength and its frequency:

𝑢 = 𝑣𝜆

Electromagnetic wave motion has the following properties.

I.The oscillating electric and magnetic fields produced by oscillating charged particles.
II.The electric and magnetic field components of an electromagnetic waves have the same
wavelength, frequency, speed and amplitude, but they are mutually perpendicular to each
other and both are perpendicular to the direction of propagation of the wave.
III.Unlike sound waves or waves produced in water, electromagnetic waves do not require
medium and can move in vacuum.

The significance of Maxwell’s theory is that it provides a mathematical description of the

general behavior of light. In particular, his model accurately describes how energy in the form of
radiation can be propagated through space as vibrating electric and magnetic fields.
Electromagnetic radiation is the emission and transmission of energy in the form of
electromagnetic waves.In a vacuum, electromagnetic waves travel at 3 × 108 m /s, which is a
physical constant, called the speed of light. This speed differs from one medium to another, but
not enough to distort our calculations significantly.

There are many types of electromagnetic radiations; EMR comes in a broad range which differs
from one another in wavelength (or frequency). These constitute what is called electromagnetic
spectrum. Different regions of the spectrum are identified by different names.

Some examples are: radio frequency region around 106 Hz, emitted by large antennas, such as
those used by broadcasting stations; microwave region around 1010 Hz used for radar; infrared
region around 1013 Hz used for heating; ultraviolet region around 1016 Hz a component of sun’s
radiation. The small portion shorter, visible light waves around 1015 Hz are produced by the
motions of electrons within atoms and molecules. is what is ordinarily called visible light. It is
only this part which our eyes can see (or detect). The shortest waves, which also have the highest
frequency, , the more energetic the radiation are associated with 𝛾(gamma) rays, which result
from changes within the nucleus of the atom . Thus, ultraviolet radiation, X rays, and 𝛾 rays are
high-energy radiation. Special instruments are required to detect non-visible radiation.

Figure 1.8 Page 17: The electromagnetic spectrum (draw)

In vacuum all types of electromagnetic radiations, regardless of wavelength, or the frequency

travel at the same speed, i.e., 3 × 108 m/s. This is called speed of light and is given the symbol
‘c‘. The frequency (v), wavelength (λ) and velocity of light (c) are related by the equation,

𝑐 = 𝑣𝜆

Problem 1 The Sheger FM Radio station in Addis Ababa, broadcasts on a frequency of 102.1M
hertz). Calculate the wavelength of the electromagnetic radiation emitted by transmitter. Which
part of the electromagnetic spectrum does it belong to?

Solution
𝑚
𝑐 3 × 108
𝑠
𝜆 =𝑣= × 𝑠 = 2.94𝑚
102.1 × 106

Problem 2.( Left for you) Calculate (a) wave number and (b) frequency of yellow radiation
having wavelength 5800 Å

1.5.2 The Quantum Theory and Photon

Some of the experimental phenomenon such as diffraction* and interference** can be explained
by the wave nature of the electromagnetic radiation. However, following are some of the
observations which could not be explained with the help of even the electromagnetic theory:
 I. the nature of emission of radiation from hot bodies (black -body radiation)
II. ejection of electrons from metal surface when radiation strikes it (photoelectric effect)
III. variation of heat capacity of solids as a function of temperature
IV. Line spectra of atoms with special reference to hydrogen.

These phenomena indicate that the system can take energy only in discrete amounts. All possible
energies cannot be taken up or radiated. It is noteworthy that the first concrete explanation for
the phenomenon of the black body radiation mentioned above was given by Max Planck.

In 1900, Max Planck, the German physicist, came to an entirely new view of matter and energy.
He made a revolutionary proposal, energy like matter is discontinuous. According to Planck,
atoms and molecules could emit or absorb energy only in discrete quantities, like small packages
or bundles. Each of these small “packets” of energy is called a quantum. The energy of a
quantum is proportional to the frequency of the radiation. The energy E of a single quantum is
given by:

𝐸 = ℎ𝜈
Where h is called Planck’s constant and ν is the frequency of radiation. The value of Planck’s
𝑐
constant is 6.63 ´ 10-34 J. s. Because ν= 𝜆 , Equation can also be expressed as
𝑐
E=h 𝜆
According to quantum theory, energy is always emitted or absorbed in integral multiples of hν;
for example, hν, 2hν, 3hν, etc. A system can transfer energy only in whole quanta. Thus, energy
seems to have particulate properties.

Problem 1 Calculate the energy (in joules) of (a) a photon with a wavelength of 5.00 × 104 𝑛𝑚
(infrared region) and (b) a photon with a wavelength of 5.00 × 10−2 nm (X ray region)..
Solution
(a) From above Equation

𝑐 3 × 108 m/s
E=h = 6.63 × 10−34 J.S × = 3.98 × 10−21 J
𝜆 5.00 × 104 × 10−9 𝑚

This is the energy of a single photon with a 5.00 × 104 𝑛𝑚 length.

(b) Following the same procedure as in (a), we can show that the energy of the photon
that has a wavelength of5.00 × 10−2 nm is

𝑐 3 × 108 m/s
E=h = 6.63 × 10−34 J.S × = 3.98 × 10−15J
𝜆 5.00 × 10−2 × 10−9 𝑚
 The Photoelectric Effect

In 1905, Albert Einstein† used the quantum theory to solve another mystery in physics, the
photoelectric effect, a phenomenon in which electrons are ejected from the surface of certain
metals exposed to light of at least a certain minimum frequency, called the threshold
frequency,νo.

These observations can be explained by assuming that EMR is quantized (consists

of photons), and the threshold frequency represents the minimum energy required to
remove the electron from the metal’s surface. Photons are particles of light or energy
packet. The minimum energy required to remove an electron is:

𝐸𝑜 = ℎ𝜈𝑜

Where Eo is the minimum energy (of the photon), and νo, the threshold frequency.

Electrons are held in a metal by attractive forces, and so removing them from the metal requires
light of a sufficiently high frequency (which corresponds to sufficiently high energy) to break
them free .The number of electrons ejected was proportional to the intensity (or brightness) of
the light, but the energies of the ejected electrons were not. Below the threshold frequency νo, or
a photon with energy less than Eo (ν < νo) no electrons were ejected no matter how intense the
light.

If a light has ν > νo, the energy in excess of that required to remove the electron is given to the
electron as kinetic energy (KE),

ℎ𝜈 = 𝐾𝐸 + 𝑊 , 𝑤 = ℎ𝑣𝑜 (W is called work function)

1
𝐾𝐸 = 𝑚𝑣 2 = ℎ𝑣 − ℎ𝑣0
2
Where KEe is the kinetic energy of an electron, m is mass of an electron, v is the velocity of an
electron, hν, is the energy of an incident photon, and hνo is the energy required to remove an
electron from the metal’s surface.

Two beams of light having the same frequency (which is greater than the threshold frequency)
but different intensities, the more intense beam of light consists of a larger number of photons;
consequently, it ejects more electrons from the metal’s surface than the weaker beam of light.
Thus, the more intense the light, the greater the number of electrons emitted by the target metal;
the higher the frequency of the light, the greater the kinetic energy of the ejected electrons.
In his theory of relativity in 1905, Einstein derived the famous equation:
𝐸 = 𝑚𝑐 2

Rearranging this equation, we have

𝐸
𝑚= 𝑐2

Where E is energy, m is mass, and c is speed of light. The main significance of the above
Equation is that energy has mass. Using this equation, we can calculate the mass associated with
a given quantity of energy or the apparent mass of a photon. For electromagnetic radiation of
wavelength, λ, the energy of each photon is given by the expression:

Then, the apparent mass of a photon of light with wavelength is given by:

𝒉𝒄
𝑬 𝒉
𝒎 = 𝟐 = ƛ𝟐 =
𝒄 𝒄 𝒄ƛ

We can summarize the important conclusions from the work of Planck and Einstein as follows: Energy is
quantized. It can occur only in discrete units called photon or quanta. EMR, which was previously
thought to exhibit only wave properties, seems to show certain characteristics of particulate matter as
well. This phenomenon is sometimes referred to as the dual nature of light and is illustrated Figure 1.9.

The dual nature of light

Example
Calculate the energy (in joules) of (a) a photon with a wavelength of 5.00 3 104 nm (infrared
region) and (b) a photon with a wavelength of 5× 104 nm (X ray region).
solution

ℎ𝑐 (6.63 × 10−34 )(3 × 108 )

𝐸= =
ƛ (5 × 104 )(1 × 10−9 )
= 3.98 × 10−21
This is the energy of a single photon with a 5 × 104 𝑛𝑚 wavelength.
(b) Following the same procedure as in (a), we can show that the energy of the photon that has a
wavelength of 5 × 10−4 𝑛𝑚 𝑖𝑠 3.98 × 10−15
The work function of cesium metal is 3.42 3 10219 J. (a) Calculate the minimum frequency of
light required to release electrons from the metal. (b) Calculate the kinetic energy of the ejected
electron if light of frequency 1.00 3 1015 s21 is used for irradiating the metal.
 Solution The minimum frequency of light needed to dislodge an electron is the point where the
kinetic energy of the ejected electron is zero. (b) Knowing both the work function and the
frequency of light, we can solve for the kinetic energy of the ejected electron.

Setting 𝑘𝑒 = 0 we write

ℎ𝑣 = 𝑊

𝑊 (3.42 × 10−19 )
𝑣= = = 5.16 × 10−34
ℎ (6.63 10−34 𝑗𝑠)

Rearranging the above gives

𝐾𝐸 = ℎ𝑣 − 𝑊

= 6.63 × 10−34 × 1 × 1015 − 3.42 × 10−19

= 3.21 × 10−19

Example

The work function of titanium metal is 6.93 × 10−19 calculate the energy of the ejected electron
if light frequency of 2.5 × 101−15 𝑠 −1 is used to irradiate a metal. What is the wavelength of a
photon (in nanometers) emitted during a transition from the n=5 state to the n= 2 state in the
hydrogen atom?

1 1
∆𝐸 = 𝑅ℎ (2 − 2)
𝑛𝑓 𝑛𝑖
1 1
∆𝐸 = 2.18 × 10−18 ( 2 − 2 )
5 2
= −4.58 × 10−19