Electromagnetic Radiation is a form of energy that travels through space in the form of waves. It consists of oscillating electric and magnetic fields that are perpendicular to each other and propagate at the speed of light. Here's a concise explanation:
Electromagnetic Radiations
Definition:
Spectrum:
Regions of the Spectrum:
Electromagnetic Spectrum
Sources:
Importance:
Properties and parameters of electromagnetic radiation describe various characteristics and attributes associated with this form of energy. Here are some key properties and parameters:
Electric and Magnetic Fields are Perpendicular
Different kinds of units are used to represent electromagnetic radiation.
Example: The speed of light in a vacuum is 3 × 108 meters per second (m/s). Calculate the frequency of light with a wavelength of 600 nanometers (nm) in a vacuum.
Solution:
To find: Frequency (ν) of light
We know that the speed of light is equal to the product of frequency and wavelength, i.e., c = νλ. Rearranging the equation, we get ν = c/λ.
Converting the wavelength from nanometers to meters:
600 nm = 600 × 10-9m
Substituting the values into the equation:
ν = (3 × 108 m/s) / (600 × 10-9m)
Simplifying the expression:
ν = 5 × 10^14 Hz
Therefore, the frequency of light with a wavelength of 600 nm in a vacuum is 5 × 10^14 hertz (Hz).
Electromagnetic Radiation is characterized based on various properties like frequency (ν ) , wavelength(λ) , time period etc. Apart from frequency and wavelength, some other parameters are also used to categorize electromagnetic radiation. One of these parameters is the wavenumber. ( ⊽).
Wave Length, Amplitude & Frequency
Polarization of Light
Understanding the properties and parameters of electromagnetic radiation is essential for fields such as optics, telecommunications, spectroscopy, and radiology. These characteristics enable us to manipulate and utilize electromagnetic radiation in numerous practical applications and scientific investigations.
Electromagnetic Radiation is light, which is present in a rainbow or a double rainbow. It also is a spectrum consisting of radio waves, microwaves, infrared waves, visible light, ultraviolet radiation, X-rays, and gamma rays. There are only two ways to transfer energy from one place to another place.
Electromagnetic radiation is a particle as well as a wave thus it is interesting to study its nature in quantum theory. Light is also electromagnetic radiation that contains frequencies.
Dual Nature of Light
A wave is a physical phenomenon characterized by its frequency, wavelength, and amplitude. In general, waves transfer energy from one location to another, in which case they have a velocity.
Here are the main points of Maxwell's electromagnetic wave theory :
1. Continuous Emission of Energy: Light is a constant release of energy from a source, known as radiant energy.
2. Oscillating Electric and Magnetic Fields: This energy consists of electric and magnetic fields that swing back and forth.
3. Perpendicular Orientation: These fields move at right angles to each other and perpendicular to the direction in which the radiation travels.
4. Electromagnetic Waves: The result is what we call electromagnetic waves or light waves.
5. Wave Characteristics: These waves exhibit characteristics of waves, like those seen in water ripples.
6. Speed of Light: They travel at an incredibly high speed, equivalent to the speed of light.
7. No Need for Material Medium: Unlike some other types of waves, these light waves can travel through empty space without needing any material substance.
In essence, Maxwell's theory helps us understand that light is a continuous flow of energy in the form of electromagnetic waves, characterized by oscillating electric and magnetic fields.
Electromagnetic Wave
Particle Nature of Light
Example 1: Calculate the no. of photons emitted by the 60-watt bulb in 10 hrs. When the light of wavelength 6000 Å is emitted by it.
= 6.5 × 1024 J
Example 2: Suppose we have a photon with a frequency of 5.0 × 1014 Hz. Calculate the energy of this photon using Planck's constant (h).
Using the equation E = hν, where ν represents the frequency, and h is Planck's constant (6.626 × 10-34 J·s), we can calculate the energy as follows:
E = (6.626 × 10-34 J·s) × (5.0 × 1014 Hz)
E ≈ 3.31 × 10-19 J
Therefore, the energy of the photon is approximately 3.31 × 10-19 Joules.
Typical Energies From Nuclear Decay:
Photoelectric effect
Emission of an electron from the metal surface when the light of suitable frequency is subjected to the metal surface. The effect is known as the photoelectric effect and the ejected electrons are known as photoelectrons.
Photoelectric Effect
1. Work function (w): It is the minimum amount of energy required to cause photoemission from the metal surface. It is also known as threshold energy or Binding energy. [Work function depends upon ionization energy and therefore w is minimum for alkali metals].
Work Function2. Threshold frequency (ν0): The minimum value of frequency that can cause photoemissions. If n < n0, then there is no photoemission.
w = h ν0 ⇒
3. Threshold wavelength (l0): The maximum value of wavelength that can cause photoemission.
If l > l0, then photoemission is not possible.
4. Intensity (I): Energy falling on the metal surface of the unit area of unit time
5. Photo intensity (IP): It is the number of photons falling per unit area per unit time.
Relation between I and Ip: I = Ip hν
Photo intensity is independent of frequency while intensity depends on frequency.
6. Power: Total energy radiated per unit time.
Example 1: A photoelectric cell emits 3.6 × 1019 electrons when illuminated with light for 2 seconds. Calculate the power of the incident light.
Solution:
Given:
Number of emitted electrons (n) = 3.6 × 1019
Time (t) = 2 seconds
Charge of an electron (e) = 1.6 × 10-19 Coulombs (C)
First, calculate the total energy (E):
E = n × e
E = (3.6 × 1019) × (1.6 × 10-19)
E = 5.76 Joules
Now, use the formula P = E/t to find the power (P):
Power = E / t
Power = 5.76 Joules / 2 seconds
Power = 2.88 Watts
Therefore, the power of the incident light on the photoelectric cell is 2.88 Watts.
Example 2: The threshold frequency for a certain metal in the photoelectric effect is 2.5 × 1014 Hz. If a light source with a frequency of 3.2 × 1014 Hz is incident on the metal, will any electrons be emitted from the metal's surface?
To answer this question, we need to compare the frequency of the incident light with the threshold frequency of the metal.
Solution:
If the frequency of the incident light (ν) is greater than or equal to the threshold frequency (ν₀), electrons will be emitted from the metal's surface. Otherwise, no electrons will be emitted.
Comparing the frequencies:
Since ν > ν₀, the frequency of the incident light is greater than the threshold frequency. Therefore, electrons will be emitted from the metal's surface.
Example: The work function (w) of a certain metal is 4.2 eV, and a photon with a wavelength of 400 nm (1 nm = 10-9 m) is incident on the metal. Calculate the maximum kinetic energy (KEmax) of the emitted electrons.
To solve this, we'll need to use the equation for the maximum kinetic energy of emitted electrons:
KEmax = hν - w
Given:
Work function (w) = 4.2 eV
Wavelength (λ) = 400 nm = 400 × 10-9 m
Planck's constant (h) = 6.626 × 10-34 J·s (Joule-seconds)
First, let's calculate the frequency (ν) using the speed of light (c) formula:
c = λν
Rearranging the formula:
ν = c / λ
The speed of light is approximately 3.00 × 108 m/s.
ν = (3.00 × 108 m/s) / (400 × 10-9 m)
≈ 7.50 × 1014 Hz
Now, we can calculate the maximum kinetic energy (KEmax):
KEmax = hν - w
= (6.626 × 10-34 J·s) × (7.50 × 1014 Hz) - (4.2 eV)
To convert eV to Joules, we'll use the conversion factor: 1 eV = 1.602 × 10-19 J.
KEmax = (6.626 × 10-34 J·s) × (7.50 × 1014 Hz) - (4.2 eV × 1.602 × 10-19 J/eV)
Calculating:
KEmax ≈ 4.97 × 10-19 J - 6.726 × 10-19 J
≈ -1.756 × 10-19 J
Note: The calculated value for KEmax is negative, which implies that no electrons will be emitted since the energy of the incident photon is lower than the work function of the metal.
It is the minimum potential required to stop the fastest moving electrons completely or it is the minimum potential at which photocurrent becomes zero.
eV0 = hν - w
eV0 = hv - hn0
It can be commented that stopping potential increases with the increase in frequency however if photo intensity is changed there is no effect on stopping potential.
Graph of Photocurrent vs cell voltage.
Example 1: An ultraviolet light of wavelength 280 nm is used in an experiment of photoelectric effect with lithium cathode (Work Function = 2.5 eV). Then calculate
(i) K.Emax (ii) Stopping potential
Q.1. Which properties among the following are false about electromagnetic waves?
(a) The energy in an electromagnetic wave is divided equally between electric and magnetic vectors.
(b) Both electric and magnetic field vectors are parallel to each other and perpendicular to the direction of propagation of the wave.
(c) These waves do not require any material medium for propagation
(d) Both electric and magnetic field vectors attain the maxima and minima at the same place and the same time
Answer: (b) The false statement is (b) "Both electric and magnetic field vectors are parallel to each other and perpendicular to the direction of propagation of the wave." In reality, these vectors are perpendicular to each other, and both are perpendicular to the direction of wave propagation.
Q.2. The ratio of the amplitude of the magnetic field to the amplitude of the electric field for electromagnetic wave propagation in a vacuum is equal to
(a) Unity
(b) Speed of light in vacuum
(c) Reciprocal of the speed of light in vacuum
(d) The ratio of magnetic permeability to electrical susceptibility in a vacuum.
Answer: (c) Reciprocal of the speed of light in vacuum
The ratio of the amplitude of the magnetic field (B0) to the amplitude of the electric field (E0) for an electromagnetic wave in a vacuum is given by the following relationship:
B0 / E0 = c
where c is the speed of light in a vacuum, approximately 3.00×108m/s. So, the correct answer is (b) "Speed of light in vacuum."
Q.3. The energy of the electromagnetic waves is of the order of 15 keV. To which part of the spectrum does it belong?
(a) X rays
(b) Infrared rays
(c) Ultraviolet rays
(d) Gamma rays
Answer: (a) X rays
Since the order of wavelength is between 10 nm and 0.001 nm, it belongs to X-rays.
Q.4. In an electromagnetic wave in free space, the root mean square value of the electric field is Erms=6V/m. The peak value of the magnetic field is
(a) 1.41 x 10-8T
(b) 2.83 x 10-8T
(c) 0.70 x 10-8T
(d) 4.23 x 10-8T
Answer: (a) 2.83 x 10-8 T
Q.5. Which of the following will emit photoelectrons when it collides with a metal?
a) UV radiation
b) Infrared radiation
c) Radio waves
d) Microwaves
Answer: Correct option will be (a) UV radiation
Reason: The photoelectric effect is the emission of electrons from a material when it is exposed to light. A sufficient frequency of light is required. It might be visible light, ultraviolet light, or X-rays. As a result, ultraviolet light causes electron emission.
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1. What is electromagnetic radiation? |
2. What are the properties and parameters of electromagnetic radiations? |
3. What is the dual behavior of electromagnetic radiation? |
4. What is the wave nature of light? |
5. What is the particle nature of light as observed in the photoelectric effect? |
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