Laser Principles:Lecturer-2

The PThe Properties of Light," light was discussed as waves. Light must be treated as waves in order to examine its propagation through space, the operation of certain optical components, and to explain phenomena such as polarization, coherence, and interference. The wave nature of light is not normally used to explain the emission and absorption of light by atoms. In these processes, light can be described as behaving more like a particle than a wave, i.e., light seems to consist of tiny entities, each with its own characteristic energy content. The "duality principle" of light states that "light cannot be described completely as either a particle or a wave, but has characteristics of both," i.e., there is a complementarily of the two concepts.
A photon is the smallest division of a light beam that retains the properties of the beam. The characteristics of a photon include its frequency, its wavelength, and its energy. A photon should not be visualized as a particle that has physical dimension or a specific location in space. More accurately, a photon can be considered a "wave packet" that has a specific energy content.
PLANCK S CONSTANT:
Energy = h*Frequency

The energy of a photon is directly proportional to the frequency of the photon and is given by Equation 1.
Equation 1
E = h?
where: E = Energy of photon in joules (J).
? = Frequency of the photon in hertz.
h = Planck s constant = 6.625 × 10–34 joule-seconds.

EXAMPLE A: CALCULATION OF THE ENERGY OF A PHOTON OF GIVEN FREQUENCY
Given: The frequency of a photon of HeNe laser light is 4.74 × 1014 Hz.
Find: The energy of the photon.
Solution: E = h?
E = (6.625 × 10–34 J.sec) (4.74 × 1014/sec)
E = 3.14 × 10–19 J
The frequency of a photon is determined by the relation described by Equation 2.
Equation 2

where: c = The speed of light in vacuum, namely 3 × 108 m/sec.
? = Wavelength in meters.
Substitution of Equation 2 into Equation 1 yields Equation 3.
Equation 3






EXAMPLE B: CALCULATION OF THE ENERGY OF A PHOTON OF A GIVEN WAVELENGTH
Given: The wavelength of a He-Ne laser light is 633 nm.
Find: The energy of the photon of this light.
Solution:
Equations 1, 2, and 3 can be used to determine the frequency, wavelength, or energy of a photon if any one of these quantities is known. Example C is a further illustration of the use of these equations.
EXAMPLE C: CALCULATION OF WAVELENGTH AND FREQUENCY OF A PHOTON OF GIVEN ENERGY
Given: A photon has an energy of 1.875 × 10–19 J.
Find: The frequency and the wavelength of the photon.
Solution: From Equation 1

From Equation 2—

ENERGY UNITS:
The unit of energy used in Planck s constant and in the preceding examples is the joule (J). The joule is the basic energy unit in the International System of Units (sometimes referred to as the MKS system—meters, kilograms, seconds) and is the amount of energy necessary to raise one kilogram mass (2.2 lb) to a height of 10.2 cm against the pull of gravity.
Another common energy unit used when talking about electron motion in atoms, energy levels in atoms, and photon energy, is the electron volt (eV). It is defined as the amount of energy acquired by an electron, when it is accelerated by a potential difference of one volt, hence the name "electron volt." One electron volt of energy is equal to 1.602? 10-19 joules of energy, that is, 1eV = 1.602? 10-19J.
The counterpart of the energy unit joule in the centimeter-gram-second system of units (the CGS system as opposed to the MKS system,) is the erg. The erg is a very small unit of energy. It takes ten million ergs to equal one joule of energy (1 erg=10-7J.) In the study of laser optics, you will rarely have to involve ergs in your calculation, but you may run across them in tables or scientific articles, so it s good to know they exist.
The energy units most commonly used to specify the energy of photons, electrons and atoms, are the joule (J) and electron volt (eV). Another unit, called a Wavenumber, is used often by spectroscopists to denote atomic and molecular energy levels. But, as the next section shows, it is not a true energy unit, even though it is useful as an indicator of energy values.
The Wavenumber of a photon is the number of wavelengths of that photon contained in a length of one centimeter. The Wavenumber is given by Equation 4. Here we use the symbol k to represent the Wavenumber in cm-1. It is perhaps even more common to use than the Greek letter ? (Nu) to represent a Wavenumber.
Equation 4

where: k = Wavenumber in cm–1
? = Wavelength in cm.
The unit of Wavenumber is the reciprocal centimeter (cm–1).

EXAMPLE D: CALCULATION OF WAVENUMBER
Given: A photon has a wavelength of 514.5 nm.
Find: The Wavenumber of the photon.
Solution: Determine the wavelength in cm:

Although the Wavenumber of a photon is not actually a measure of the energy of the photon, it is proportional to the energy and often is utilized to express the energy of a photon or the energy level within an atom.
CONVERSION OF ENERGY UNITS:
Table 1 lists the conversion factors for converting a quantity of energy expressed in one unit to an equivalent expression in another unit. The table features only the most commonly used energy units—the joule, cm–1 and eV. To use this table, locate the original energy unit in a row along the left column of the table, then move horizontally along the row to the right to the column headed by the desired energy unit. The factor located in the box at the intersection of horizontal row and vertical column is the desired conversion factor. Multiplication of the original expression by the conversion factor listed yields the equivalent energy expressed in the desired unit. Examples E, F and G illustrates this procedure. Each of these Examples indicates the importance of tracking units carefully (doing a "dimensional analysis") and using the correct conversion factors.
TABLE 1. CONVERSION FACTORS FOR ENERGY UNITS
Unit joule electron
volt cm–1
joule 1 6.24 × 1018 5.034 × 1022
electron
volt 1.602 × 10–19 1 8065.73
cm–1 1.9864 × 10–23 1.24 × 10–4 1

EXAMPLE E: CONVERTING ENERGY IN JOULES TO CM-1
Given: A He-Ne laser photon has a wavelength of 632.8 nanometers
Find: (a) Photon energy in joules
(b) Photon energy in cm–1
Solution: (a) With Equation 3, find energy of He-Ne laser photon in joules.
Where h=6.625? 10-34J• sec
c=3? 108m/sec
? =632.8nm=632.8? 10-9m
Use Table 1 to convert 3.14? 10-19 joules to cm-1. Locate "joule" in the first row in the left hand column. Follow this row over to the column headed "cm-1." At the intersection of the row and column, find the conversion factor 5.034? 1022 cm-1/joule. Multiply this factor by 3.14? 10-19 joules to change the energy from joules to cm-1.
The "joule" units cancel, and you get E=15,806.8cm-1



EXAMPLE F: CONVERTING ENERGY IN CM-1 TO JOULES
Given: In energy notation involving the unit cm–1, the energy of a Nd:YAG solid state laser photon is given as 9398.5cm–1
Find: (a) The energy of the photon in joules
(b) The wavelength of the photon
Solution: (a) Using Table 1, as in Example E, the conversion factor from cm-1 to joules is found to be 1.9864? 10-23 joules/cm-1. Multiplying the value 9398.5cm-1 by this conversion factor to change cm-1 to joules gives the photon energy in joules.
After cancellation of the cm-1 units, the result is E=1.87? 10-19joules.
(b) The wavelength (in meters) is obtained from the equation E=hc/? which when solved for the wavelength ? gives ? hc/E
where
h=6.625? 10-34J• sec
c=3? 108m/sec
E=1.87? 10-19J
So, ? = 1.063? m










EXAMPLE G: CONVERTING ENERGY IN JOULES TO eV
Given: A Nd:YAG laser photon of energy 1.87? 10-19joules
Find: Photon energy in electron volts (eV)
Solution: Solution: Using Table 7.1, as indicated in Examples E and F, locate "joule" in the left column and follow its row over to the column labeled "electron volt." At the intersection of the row and column, find the conversion factor 6.24? 1018 eV/joule. Multiply the given energy in joules by this factor to get the desired energy in electron volts.
E=(eV) = (1.87? 10-19J)(6.24? 1018eV/J)
E=1.17eV