Electromagnetic Waves Physics Knowledge Assessment: 30 MCQs

Explanation:

To find the wavelength of the electromagnetic wave in the medium, we follow these steps:

1. Determine the speed of the wave in the medium:

The speed of light c in a vacuum is 3 × 10⁸ m/s. The speed v of the wave in the medium is given by:

v = c / n​

where n is the refractive index.

v = 3 × 10⁸ m/s / 1.5 = 2 × 10⁸ m/s

2. Calculate the wavelength using the wave equation:

The wavelength λ in the medium is given by:

λ = v / f​

where f is the frequency of the wave.

λ = 2 × 10⁸ m/s / 8 × 10¹⁴ Hz = 2.5 × 10⁻⁷ m

3. Convert the wavelength to angstroms (Å):

λ = 2.5 × 10⁻⁷ m × 10¹⁰ A˚/m = 2500 A˚

Conclusion:

The wavelength of the electromagnetic wave with a frequency of 8×10¹⁴ in a medium with a refractive index of 1.5 is:

(b) 2500 Å

Explanation:

Given:

• Energy of the photon: E = 10 eV

The relationship between energy (E) and wavelength (λ) for electromagnetic radiation is expressed as: E = hc / λ​

Where:

• Planck's constant (h) is 4.135667696 × 10⁻¹⁵ eV⋅s
• Speed of light (c) is 3 × 10⁸ m/s

The electromagnetic spectrum regions are defined as follows:

• Ultraviolet (UV) rays: approximately 3 eV to 124 eV
• Infrared (IR) rays: approximately 0.001 eV to 1.7 eV
• Microwaves: approximately 10⁻⁵ eV to 10⁻³ eV
• X-rays: approximately 124 eV to 10⁶ eV

Calculation:

To determine the wavelength (λ) corresponding to a 10 eV photon:

λ = hc / E

λ = (4.135667696 × 10⁻¹⁵ eV⋅s) × (3 × 10⁸ m/s)​ / 10eV

λ = 1.2407003088 × 10⁻⁷ m

λ = 124 nm

Given that the wavelength of 124 nm falls within the ultraviolet range, the correct answer is:

(a) Ultraviolet rays

Explanation:

The frequency f is 6 × 10¹⁴ Hz.

The wavelength λ is calculated as:

λ = c / f

= 3 × 10⁸ m/s / 6 × 10¹⁴ Hz

= 5 × 10⁻⁷ m.

Converting meters to angstroms:

5 × 10⁻⁷ m

= 5000 A˚

The correct option is: (d) 5000 A˚

Explanation:

Speed of light:

c = 3 × 10⁸ m/s

Wavelength:

λ = 2A˚

= 2 × 10⁻¹⁰ m

The frequency f is calculated using the formula:

f = c / λ.

The frequency f is given by:

f = c / λ

= 3×10⁸ m/s / 2×10⁻¹⁰ m

​= 1.5×10¹⁸ Hz​

Explanation:

Given:

Speed of light:

c = 3 × 10⁸ m/s

Wavelength:

λ = 1 A˚

= 1 × 10⁻¹⁰ m

The frequency f of the photon is calculated using the formula:

f = c / λ

Substituting the given values:

f = (3 × 10⁸ m/s) / (1 × 10⁻¹⁰ m) = 3 × 10¹⁸ Hz

Converting the frequency to different units:

1. In hertz (Hz): 3 × 10¹⁸Hz
2. In gigahertz (GHz): (3 × 10¹⁸Hz) × (1 GHz / 10⁹ Hz) = 3 × 10⁹ GHz
3. In megahertz (MHz): (3 × 10¹⁸Hz) × (1 MHz / 10⁶ Hz) = 3 × 10¹² MHz

Thus, the correct answer is: (d) All of the above

Explanation:

Background Information:

Photon Energy:

• The energy of a photon is given by E = hc / λ​
• Planck's constant h = 6.626 × 10⁻³⁴J\cdotps
• Speed of light c = 3 × 10⁸m/s
• Wavelength λ = 1 A˚ = 1 × 10⁻¹⁰m

Relativistic Mass:

• The energy-mass equivalence principle is E = mc²
• The effective mass mmm can be derived from the photon's energy using m = E / c²

First, calculate the energy of the photon:

E = (6.626 × 10⁻³⁴ J\cdotps) × (3 × 10⁸ m/s) / 1 × 10⁻¹⁰ m

E = 1.9878 × 10⁻²⁵ J / 10⁻¹⁰

E = 1.9878 × 10⁻¹⁵ J

Now, convert this energy to relativistic mass:

m = E / c²  

m = 1.9878 × 10⁻¹⁵ J / (3 × 10⁸ m/s)²

m = 1.9878 × 10⁻¹⁵ / 9 × 10¹⁶

m ≈ 2.2 × 10⁻³² kg

Given this calculation, the effective mass of the photon with a wavelength of 1 A˚ is approximately 2.2 × 10⁻³² kg.

Therefore, the correct answer is: (b) 10⁻³² kg

Explanation:

The correct relationship between the speed of light in a vacuum (c), the absolute permittivity of vacuum (ε₀​), and the absolute permeability of vacuum (μ₀​) is given by:

This formula is derived from Maxwell's equations and accurately describes the propagation speed of electromagnetic waves in a vacuum. The speed of light c is approximately 3 × 10⁸ meters per second, and this relationship reflects the fundamental electromagnetic properties of free space.

Therefore, the correct answer is: (b)      

Explanation:

Given:

Planck's constant:

h = 6.626 × 10⁻³⁴ Js

Energy of the X-ray photon:

E = 3.31 × 10⁻¹⁶ J

The frequency f of the photon is calculated using the formula:

E = hf

Solving for f:

f  = E / h

Substituting the given values:

f = 3.31 × 10⁻¹⁶ J / 6.626 × 10⁻³⁴ Js

f = 4.995 × 10¹⁷ Hz

Therefore, the correct answer is: (a) 5 × 10¹⁷ Hz

Explanation:

Background Information:

Photon Energy:

• The energy E of a photon is given by E = hc / λ, where h is Planck's constant and c is the speed of light.
• Higher frequency (or lower wavelength) photons have higher energy.

Relativistic Mass:

The relativistic mass m of a photon is derived from its energy using E = mc²

Energy Calculation:

• For X-ray photons with λ = 1 A˚ = 1 × 10⁻¹⁰m:

E_X-ray = (6.626 × 10⁻³⁴ J\cdotps) × (3 × 10⁸ m/s) / 1 × 10⁻¹⁰ m

E_X-ray ​= 1.9878 × 10⁻¹⁵ J

• For typical UV photons with λ = 200 nm = 2 × 10⁻⁷m:

E_UV = (6.626 × 10⁻³⁴ J\cdotps) × (3 × 10⁸ m/s) / 2 × 10⁻⁷ m​

E_UV = 9.939 × 10⁻¹⁹ J

Relativistic Mass Calculation:

Relativistic mass mmm is given by m = E / c².

For X-ray photons:

m_X-ray = 1.9878 × 10⁻¹⁵ J / (3 × 10⁸ m/s)²

m_X-ray = 2.208 × 10⁻³² kg

For UV photons:

m_UV = 9.939 × 10⁻¹⁹ J / (3×10⁸ m/s)²

m_UV​ = 1.104 × 10⁻³⁶ kg

Conclusion:

The relativistic mass of X-ray photons is significantly higher than that of UV photons due to their higher energy.

Therefore, the correct answer is:

(a) X-ray photons with a wavelength of 1 A˚ are heavier than ultraviolet photons

Explanation:

Given:

Planck's constant:

h = 6.626 × 10⁻³⁴ Js

Speed of light:

c = 3 × 10⁸ m/s

Conversion of units:

1 A˚ = 1 × 10⁻¹⁰ m

0.01 mm = 1 × 10⁻⁵ m

The energy E of a photon is calculated using the formula:

E = hc / λ ​

Calculate the energy for each wavelength and determine the maximum:

1. For 1 A˚ = 1 × 10⁻¹⁰m:

E = (6.626 × 10⁻³⁴ Js) × (3 × 10⁸ m/s) / 1 × 10⁻¹⁰ m = 1.9878×10⁻¹⁵ J

2. For 5896 A˚ = 5896 × 10⁻¹⁰m:

E = (6.626 × 10⁻³⁴ Js) × (3 × 10⁸ m/s) / 5896 × 10⁻¹⁰ m = 3.373 × 10⁻¹⁹ J

3. For 1122 A˚ = 1122 × 10⁻¹⁰m:

E = (6.626 × 10⁻³⁴ Js) × (3 × 10⁸ m/s) / 1122 × 10⁻¹⁰ m = 1.772 × 10⁻¹⁸ J

4. For 0.01 mm = 1 × 10⁻⁵m:

E = (6.626 × 10⁻³⁴ Js) × (3 × 10⁸ m/s) / 1 × 10⁻⁵ m = 1.9878 × 10⁻²⁰ J

The photon with the wavelength of 1 A˚ has the highest energy among the given options, calculated as 1.9878 × 10⁻¹⁵ J. Comparing the calculated energies for all wavelengths, 1 A˚ results in the maximum energy. The correct answer is: (a) 1 A˚

Explanation:

Background Information:

• Photons are massless particles with zero rest mass.
• Their energy E is given by E = hν, where h is Planck's constant and ν is the frequency.
• X-ray photons have higher energy (and thus higher frequency) compared to ultraviolet photons.
• The relativistic mass of a photon can be inferred from its energy via E = mc²

Since X-ray photons have higher energy than ultraviolet photons, they effectively have a greater relativistic mass when considering the energy-mass equivalence principle. Therefore, in the context of their energy (and thus their effective relativistic mass), X-ray photons can be considered "heavier" than ultraviolet photons.

So, the correct answer is: (a) Heavier than ultraviolet ray photons

Explanation:

• While photons have zero rest mass, their energy is proportional to their effective relativistic mass.
• X-ray photons have more energy compared to ultraviolet photons, and hence, they can be considered heavier in this specific context.

The correct answer is: (a) Heavier than ultraviolet ray photons

Explanation:

Given frequency:

f = 1.55 × 10⁶ Hz

Speed of light:

c = 3 × 10⁸ m/s

Wavelength calculation:

λ = c / f = (3 × 10⁸ m/s) / (1.55 × 10⁶ Hz) ≈ 193.55 m

Based on this calculation, the closest option to the actual wavelength is 200 m.

Explanation:

Given:

Planck's constant:

h = 6.626 × 10⁻³⁴ Js

The relation E = hν is used to describe the energy of photons in electromagnetic radiation.

Consider the following options and determine for which the energy-frequency relation is not valid:

(a) X-rays: High-energy electromagnetic waves

(b) Ultraviolet radiation: Electromagnetic waves with shorter wavelengths than visible light

(c) Cathode rays: Streams of electrons emitted from the cathode in a vacuum tube

(d) Microwaves: Electromagnetic waves with longer wavelengths than infrared radiation

The energy-frequency relation E = hν applies to electromagnetic waves (options (a), (b), and (d)). However, cathode rays are streams of electrons, and their energy is not described by E = hν. Thus, the correct answer is: (c) Cathode rays

Explanation:

Given:

Planck's constant;

h = 6.626 × 10⁻³⁴ Js

Speed of light:

c = 3 × 10⁸ m/s

Wavelength:

λ = 1 × 10⁻¹⁰ m

The energy E of a photon is calculated using the formula: E = hc / λ

Substituting the given values:

E = (6.626 × 10⁻³⁴ Js) × (3 × 10⁸ m/s) / (1×10⁻¹⁰ m)

E = 1.9878 × 10^-25 Js m​ / 1 × 10⁻¹⁰ m

E = 1.9878 × 10^-15 Js m

So, the energy of a photon with a wavelength of 1 A˚ is indeed approximately 1.99 × 10⁻¹⁵ J. Thus, the correct answer is: (c) 10⁻¹⁵ J

Explanation:

Background Information:

• Photons are elementary particles that serve as the quantum of the electromagnetic field.
• Phonons are quanta of lattice vibrations in crystalline solids and are not related to gravitational fields.
• Hyperons are baryons containing one or more strange quarks, which are not associated with gravitational field quanta.
• Muons are elementary particles similar to electrons but with a greater mass and are not related to gravitational field quanta.
• Gravitons are hypothetical particles proposed to mediate the force of gravity in quantum field theory.

Explanation:

• The graviton is the theoretical quantum particle that mediates the gravitational force, similar to how the photon mediates the electromagnetic force.
• While the existence of gravitons has not been experimentally confirmed, they are predicted by theories attempting to unify quantum mechanics and general relativity.

The correct answer is: (b) Graviton

Explanation:

Given:

Planck's constant:

h = 6.626 × 10⁻³⁴ Js

Speed of light:

c = 3 × 10⁸ m/s

Wavelength:

λ = 0.5 A˚ = 0.5 × 10⁻¹⁰ m

Conversion factor:

1 eV = 1.602 × 10⁻¹⁹ J

The energy E of a photon is calculated using the formula:

E = hc / λ

Converting the result to electron volts:

E(eV) = (hc / λ) × (1 / 1.602 ×10⁻¹⁹) ​

First, calculate hc / λ​:

E = (6.626 × 10⁻³⁴ Js) × (3 × 10⁸ m/s) / (0.5 × 10⁻¹⁰ m)

E = 1.9878 × 10⁻²⁵ Js m / 0.5 × 10⁻¹⁰ m

E = 3.9756 × 10⁻¹⁵ J

Convert the energy to electron volts:

E(eV) = 3.9756 × 10⁻¹⁵ J / 1.602 × 10⁻¹⁹ J/eV

E(eV) = 24800 eV

Therefore, the correct answer is: (b) 24800 eV