10th Class Physics Notes 2026 — Chapter 20 – Federal Board

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Class 10 Physics – Chapter 20: Electromagnetic Waves

Board: Federal Board (FBISE)

Multiple Choice Questions

1. Identify the electromagnetic wave with the greatest ability to penetrate matter.

A. Radio waves
B. Ultraviolet
C. X-rays
D. Gamma rays

Correct Answer: (D) Gamma rays

2. Which type of radiation primarily carries heat energy?

A. Radio waves
B. Ultraviolet
C. Infrared
D. Visible light

Correct Answer: (C) Infrared

3. A wavelength of 10 cm corresponds to which part of the electromagnetic spectrum?

A. Microwaves
B. Ultraviolet
C. Infrared
D. X-rays

Correct Answer: (A) Microwaves

4. The atomic structure of solids or cracks in solid structures can be investigated by:

A. Microwaves
B. Ultraviolet
C. Gamma rays
D. X-rays

Correct Answer: (D) X-rays

5. Which waves are used for treating muscle aches in physiotherapy?

A. Microwaves
B. Ultraviolet
C. Infrared
D. X-rays

Correct Answer: (C) Infrared

6. X-rays have higher energy than visible light because of their:

A. Shorter wavelength
B. Longer wavelength
C. Lower frequency
D. Higher speed

Correct Answer: (A) Shorter wavelength

7. Electromagnetic waves are produced due to the motion of which of the following?

A. Neutral particles
B. Charged particles
C. Accelerating neutral particles
D. Accelerating charged particles

Correct Answer: (D) Accelerating charged particles

8. If the frequency of an electromagnetic wave is \(2\times10^{18}\text{ Hz}\), to which part of the spectrum does it belong?

A. Microwaves
B. Ultraviolet
C. Infrared
D. X-rays

Correct Answer: (D) X-rays

9. Which colour has the longest wavelength in the visible spectrum?

A. Green
B. Violet
C. Orange
D. Red

Correct Answer: (D) Red

10. Electromagnetic waves exhibit particle-like behaviour in quantum mechanics. These particles are called:

A. Atoms
B. X-rays
C. Electrons
D. Photons

Correct Answer: (D) Photons

11. The energy of an electromagnetic wave depends directly on its:

A. Wavelength
B. Amplitude
C. Frequency
D. Speed

Correct Answer: (C) Frequency

12. When an electromagnetic wave enters from one medium into another, which quantity does not change?

A. Wavelength
B. Speed
C. Frequency
D. Energy

Correct Answer: (C) Frequency

13. On halving the area of a totally reflecting surface, the radiation pressure will:

A. Remain the same
B. Double
C. Halve
D. Quadruple

Correct Answer: (A) Remain the same

14. Solar sails are preferred for deep-space missions because they require:

A. Electric energy
B. Fuel
C. Chemical energy
D. No fuel

Correct Answer: (D) No fuel

15. Which of these electromagnetic waves has the least momentum?

A. Gamma rays
B. Ultraviolet
C. Infrared
D. Radio waves

Correct Answer: (D) Radio waves

Constructed Response Questions

1. What colour does the solution appear when viewed from above? Why?

Answer: When observed from the side or above, the colloidal sulfur solution appears blue.

This occurs due to Rayleigh scattering, in which fine colloidal sulfur particles selectively scatter shorter wavelengths of light (blue light) more effectively than longer wavelengths when white light passes through the mixture.

2. What colour spot appears on the screen ‘N-M’ after the light passes through the whole beaker?

Answer: The spot appearing on the screen ‘N-M’ will be orange-red (or red).

As white light travels through the beaker containing colloidal sulfur, the shorter wavelengths (blue light) are scattered sideways. Consequently, the transmitted light reaching the screen contains a greater proportion of the longer, less-scattered wavelengths, such as red and orange light.

3. What phenomenon is observed in this activity?

Answer: The phenomenon demonstrated in this activity is the Tyndall effect, along with Rayleigh scattering of light by colloidal particles.

Short Response Questions

1. A leaking microwave oven in a home can sometimes cause interference with the home’s Wi-Fi system. Why?

Both microwave ovens and standard Wi-Fi routers operate in the same unlicensed frequency band of 2.4 GHz.

A leaking microwave oven releases radiation at this frequency, creating electromagnetic noise that overlaps with and disrupts Wi-Fi signal transmission.

2. Give an example of energy carried by an electromagnetic wave.

Sunlight carrying radiant energy that warms the Earth’s surface and drives photosynthesis is a common example of electromagnetic energy.

Solar panels also absorb electromagnetic radiation from the Sun and directly convert it into usable electrical energy.

3. Your friend says that more patterns and colours can be seen on the wings of birds when viewed in ultraviolet light. Would you agree with your friend? Explain.

Yes, I agree with my friend.

Many bird feathers contain specialised structures and pigments that reflect ultraviolet radiation, creating vivid UV patterns that are invisible to the human eye.

Birds possess UV-sensitive photoreceptors in their eyes, allowing them to see these distinct patterns and colours.

4. Can the human body detect electromagnetic radiations that are outside the visible region of the spectrum?

Yes, the human body can detect some effects of non-visible electromagnetic radiation through sensory receptors and physiological reactions.

Thermal receptors in the skin detect infrared radiation as heat, while skin tissue responds to ultraviolet radiation by producing melanin, which can lead to tanning or sunburn.

5. What property of light corresponds to loudness in sound?

The intensity of light, which is related to its amplitude, corresponds to the loudness of sound.

Just as a higher wave amplitude in sound represents greater acoustic energy and loudness, a higher wave amplitude in light represents greater intensity and perceived brightness.

6. Certain orientations of a television antenna give better reception than others for a particular station. Explain.

Television broadcast waves are polarised, meaning their electric field oscillates in a specific plane.

Antenna reception is maximised when the receiving antenna rods are aligned parallel to the plane of polarisation of the incoming electromagnetic wave.

7. Why are microwaves used for satellite TV and radio waves used for terrestrial TV?

Microwaves have high frequencies that allow them to pass through the Earth’s ionosphere and reach orbiting satellites.

Conversely, longer radio waves can bend around surface obstacles and may reflect from atmospheric layers, making them useful for terrestrial broadcasting.

8. Can we use X-rays and gamma rays for broadcasting radio and TV signals?

No, X-rays and gamma rays are not suitable for broadcasting radio and TV signals because their extremely high energy makes them ionising and biologically hazardous.

Furthermore, they pass through ordinary electronic antennas without interacting in the same way as lower-frequency waves and are difficult to use for conventional communication systems.

9. When you stand outdoors in sunlight, why can you feel the energy that sunlight carries but not the momentum it carries?

Sunlight transfers noticeable thermal energy through photon absorption in skin tissue, causing a sensation of heat.

However, because the speed of light \(c\) is extremely large, the radiation pressure and momentum transferred by photons are very small and below the threshold of human touch.

10. A newscaster in the studio speaks to a reporter at a remote place, but the reporter’s voice is sometimes delayed. What causes this delay?

The delay occurs because the audio signal must be encoded, transmitted to a geostationary satellite, relayed back to Earth, and processed electronically.

The signal travels tens of thousands of kilometres at the finite speed of light. This travelling time, along with digital encoding and processing delays, creates a noticeable time lag.

11. Is the visible region a major portion of the electromagnetic spectrum? Explain.

No, the visible region forms an extremely small fraction of the vast electromagnetic spectrum.

It spans a narrow wavelength range of approximately 400 nm to 700 nm, lying between the much broader ultraviolet and infrared regions.

12. As light passes through the atmosphere, it scatters. Does the same happen with infrared light and radio waves? Explain.

Rayleigh scattering is inversely proportional to the fourth power of wavelength:

\[ I\propto\frac{1}{\lambda^4} \]

Infrared light and radio waves have significantly longer wavelengths than visible light. Therefore, their Rayleigh scattering in the atmosphere is much weaker, allowing them to pass through with comparatively little scattering.

13. Why is the radiation pressure of an electromagnetic wave greater when it is perfectly reflected than when it is absorbed by a surface?

Radiation pressure depends on the change in momentum of photons striking a surface.

When photons are completely absorbed, their momentum changes by:

\[ \Delta p=p \]

When photons are totally reflected, they reverse their direction. Therefore, their momentum change is:

\[ \Delta p=2p \]

Thus, the momentum transferred to a perfectly reflecting surface is twice that transferred to a perfectly absorbing surface. Consequently, the radiation pressure is greater for reflection.

14. How are photons produced?

Photons are produced when bound electrons transition from higher atomic energy levels to lower energy levels, releasing the energy difference as a quantum of light.

They can also be emitted when charged particles accelerate or oscillate, and during high-energy nuclear reactions.

15. Explain why clouds appear white in colour.

Clouds contain large water droplets and ice crystals that are much larger than the wavelengths of visible light.

These droplets undergo Mie scattering, which scatters the different visible wavelengths, such as red, green, and blue, nearly equally. The combined scattered light reaching our eyes makes clouds appear white.

Long Response Questions

1. Break down the electromagnetic spectrum in terms of wavelength and frequency. Why is the speed of electromagnetic waves in air nearly the same as in a vacuum?

Electromagnetic Spectrum Breakdown:

The electromagnetic spectrum is the continuous range of electromagnetic radiation categorised according to frequency (\(f\)) and wavelength (\(\lambda\)). These quantities are related by:

\[ c=f\lambda \]

The electromagnetic spectrum is arranged from the lowest frequency (longest wavelength) to the highest frequency (shortest wavelength) as follows:

1. Radio Waves

  • Wavelength: \(\lambda>1\text{ m}\)

  • Frequency: \(f<300\text{ MHz}\)

  • Uses: Broadcasting and telecommunications.

2. Microwaves

  • Wavelength: \(\lambda=1\text{ mm to }1\text{ m}\)

  • Frequency: \(f=300\text{ MHz to }300\text{ GHz}\)

  • Uses: Radar and cooking.

3. Infrared (IR)

  • Wavelength: \(\lambda=700\text{ nm to }1\text{ mm}\)

  • Frequency: \(f=300\text{ GHz to }430\text{ THz}\)

  • Uses: Thermal radiation and heat detection.

4. Visible Light

  • Wavelength: \(\lambda=400\text{ nm to }700\text{ nm}\)

  • Frequency: \(f=430\text{ THz to }750\text{ THz}\)

  • Description: The part of the spectrum detectable by the human eye.

5. Ultraviolet (UV)

  • Wavelength: \(\lambda=10\text{ nm to }400\text{ nm}\)

  • Frequency: \(f=750\text{ THz to }30\text{ PHz}\)

  • Uses: Sterilisation and other applications.

6. X-rays

  • Wavelength: \(\lambda=0.01\text{ nm to }10\text{ nm}\)

  • Frequency: \(f=30\text{ PHz to }30\text{ EHz}\)

  • Uses: Medical imaging and investigating crystal structures.

7. Gamma Rays

  • Wavelength: \(\lambda<0.01\text{ nm}\)

  • Frequency: \(f>30\text{ EHz}\)

  • Sources: Nuclear reactions and cosmic events.

Speed of Electromagnetic Waves in Air Compared with Vacuum:

In a vacuum, electromagnetic waves travel at the speed of light:

\[ c=\frac{1}{\sqrt{\mu_0\varepsilon_0}} \]
\[ c\approx3\times10^8\text{ m/s} \]

Air is an extremely dilute gaseous medium with a refractive index of approximately:

\[ n\approx1.000293 \]

The permittivity and permeability of air are very close to those of a vacuum. Therefore, light slows down by less than 0.03% when travelling through air.

As a result, the speed of electromagnetic waves in air is practically equal to their speed in a vacuum.

2. Evaluate the importance of radio waves in communication systems and astronomy compared with other electromagnetic waves. Why are they uniquely suited for this purpose?

Importance in Communication and Astronomy:

Radio waves form the backbone of global telecommunications. They are used for long-distance television, radio, cellular networks, and satellite transmissions.

In radio astronomy, radio waves allow astronomers to observe regions of space hidden by thick cosmic dust clouds that block visible light. They help reveal distant pulsars, quasars, cosmic microwave background radiation, and the structure of galactic arms.

Why Radio Waves Are Uniquely Suited:

1. Long Wavelength and Diffraction: Their long wavelengths allow them to diffract around large obstacles, such as hills and buildings, and follow the curvature of the Earth.

2. Atmospheric Transmission Window: Earth’s atmosphere is transparent to many radio frequencies, allowing ground-based observatories to receive signals from space.

3. Ionospheric Reflection: Some low-frequency radio waves reflect from the ionosphere, permitting over-the-horizon communication without requiring satellites.

3. Compare the use of microwaves in household applications with their industrial or scientific uses. How do their characteristics make them versatile?

Comparison of Uses:

Parameter

Household Applications

Industrial / Scientific Uses

Primary Function

Dielectric heating of food in microwave ovens.

Radar tracking, satellite links, materials synthesis, and plasma heating.

Frequency Band

Typically 2.45 GHz.

Ranges from 1 GHz to 300 GHz.

Mechanism

Rotates polar water molecules in food to generate heat.

High-resolution spatial reflections, molecular spectroscopy, and particle acceleration.

Characteristics Explaining Their Versatility:

  • Selective Absorption: Microwaves heat polar molecules, such as water, directly without significantly heating the surrounding air.

  • Penetration Depth: They penetrate several centimetres into organic materials, allowing rapid internal heating.

  • High Line-of-Sight Directivity: Their shorter wavelengths compared with ordinary radio waves allow narrow beam formation for precision radar and satellite communication.

4. What are infrared radiations, how can they be produced, and what are their characteristics?

Definition:

Infrared (IR) radiation consists of electromagnetic waves with wavelengths ranging from 700 nm to 1 mm. It lies between visible light and microwaves in the electromagnetic spectrum.

Production:

Infrared radiation is produced by the thermal motion of molecules and atoms.

Any object with a temperature above absolute zero (0 K) emits infrared radiation due to the vibrational and rotational motion of its constituent molecules.

Characteristics:

  1. Thermal Effect: Infrared radiation is readily absorbed by matter, causing molecular vibrations that are experienced as heat.

  2. Penetration Through Haze: It scatters less than visible light, allowing infrared imaging to work in conditions involving smoke, dust, and atmospheric fog.

  3. Lower Photon Energy: It has lower energy than visible light:

\[ E=hf \]

Infrared radiation is non-ionising.

5. What is visible light? Why is visible light crucial for life on Earth?

Definition:

Visible light is the narrow segment of the electromagnetic spectrum, approximately 400 nm to 700 nm, that can be detected by the human eye.

Importance for Life on Earth:

1. Photosynthesis: Plants use visible light wavelengths, primarily red and blue, to drive photosynthesis. They convert solar energy into chemical energy in the form of glucose and release oxygen (\(O_2\)).

2. Vision and Navigation: Visible light provides the primary medium for animal vision. It enables navigation, foraging, predator detection, and communication.

3. Thermal Regulation and Climate: Visible sunlight contributes to maintaining global temperatures suitable for liquid water and biological activity.

6. Ultraviolet radiation has both beneficial and harmful effects. Discuss its characteristics, applications, and risks. How can we balance its advantages, such as sterilisation, with its dangers, such as skin cancer?

Characteristics and Applications:

Ultraviolet (UV) radiation has wavelengths ranging from 10 nm to 400 nm. It possesses higher energy than visible light.

Beneficial Applications:

  • Vitamin D Production: UV radiation helps trigger vitamin D synthesis in human skin.

  • Sterilisation: UV-C radiation is used for water and air sterilisation because it can destroy microorganisms.

  • Forensic Detection: UV radiation is used to detect certain substances and materials.

  • Industrial Applications: It is used in polymerisation and photolithography.

Risks:

High-energy UV photons, particularly UV-B and UV-C, can damage DNA and break chemical bonds. Excessive exposure can cause:

  • DNA damage and cellular mutations.

  • Skin cancer, including melanoma.

  • Cataracts.

  • Premature skin ageing.

Balancing Advantages and Dangers:

Controlled applications using enclosed UV chambers for air and water purification can prevent direct human exposure.

For outdoor safety, people should:

  • Apply broad-spectrum sunscreen.

  • Wear UV-blocking eyewear.

  • Avoid excessive exposure during peak sunlight hours.

7. How are X-rays different from the rest of electromagnetic radiations? Discuss in detail.

Distinctive Origin and High Energy:

Unlike many optical and radio waves, which can be produced by electron transitions or oscillating charges, X-rays are generated by high-energy atomic processes.

They are mainly produced through:

  • Bremsstrahlung Radiation: When high-speed electrons are rapidly decelerated.

  • Inner-Shell Electron Transitions: When electrons move between inner atomic energy levels in heavy atoms.

Unique Characteristics:

1. High Penetrating Power: X-rays have high energy, typically ranging from 100 eV to 100 keV. They can pass through soft biological tissues while being absorbed more strongly by denser structures such as bones.

2. Ionisation Ability: X-rays possess sufficient photon energy to remove bound electrons from atoms, causing ionisation.

3. Atomic-Scale Wavelength: Their wavelengths, approximately 0.01 nm to 10 nm, are comparable to atomic lattice spacings. This makes them useful in X-ray crystallography for analysing crystal structures.

8. Gamma rays are the most energetic electromagnetic waves. How do their properties make them suitable for applications in nuclear science and astrophysics?

Properties:

Gamma rays have the highest frequency and shortest wavelength in the electromagnetic spectrum.

\[ f>10^{19}\text{ Hz} \]
\[ \lambda<10^{-11}\text{ m} \]

They carry extremely high photon energy, typically greater than 100 keV.

Applications in Nuclear Science and Astrophysics:

  • Nuclear Physics: Gamma rays are emitted during radioactive decay and nuclear fission and fusion reactions. Their precise energy spectra allow nuclear physicists to study nuclear energy levels.

  • Cancer Radiotherapy: Concentrated gamma-ray beams, such as those produced by Cobalt-60 sources, are used to destroy targeted malignant tumour cells.

  • Astrophysics: High-energy cosmic sources, including supernovae, black hole accretion disks, and gamma-ray bursts, emit gamma rays. Their study helps astrophysicists investigate extreme high-energy phenomena in the universe.

9. What potential hazards are associated with different parts of the electromagnetic spectrum? Compare the risks posed by ionising and non-ionising radiation, and explain why ionising radiation is more dangerous to living tissues and biological systems.

Spectrum Hazards and Comparison of Ionising and Non-Ionising Radiation:

1. Non-Ionising Radiation (Radio Waves, Microwaves, Infrared, and Visible Light)

Non-ionising radiation does not normally have enough photon energy to remove electrons from atoms. However, at high power levels, it can cause heating of body tissues, such as microwave burns.

2. Ionising Radiation (Ultraviolet, X-rays, and Gamma Rays)

Ionising radiation has photons with sufficient energy to remove electrons from atoms and molecules. This can damage biological tissues.

Why Ionising Radiation Is More Dangerous:

Ionising radiation can eject electrons directly from biological molecules, including DNA, or cause the ionisation of cellular water, producing free radicals such as hydroxyl radicals (\(OH\cdot\)).

These reactive species can cause:

  • Single-strand and double-strand DNA breaks.

  • Cell death.

  • Genetic mutations.

  • Increased risk of cancer.

Therefore, ionising radiation is more dangerous because it can directly damage the molecules essential for normal biological functions.

10. Explain the scattering of light in the atmosphere and why the colour of the sky appears different at different times of the day.

Rayleigh Scattering Mechanism:

Air molecules, mainly nitrogen (\(N_2\)) and oxygen (\(O_2\)), are much smaller than the wavelengths of visible light.

According to Rayleigh’s Law, the intensity of scattering is inversely proportional to the fourth power of wavelength:

\[ I\propto\frac{1}{\lambda^4} \]

Daytime Sky:

At midday, the Sun is nearly overhead, and sunlight travels a relatively short distance through the atmosphere.

Shorter blue wavelengths (approximately 400–450 nm) scatter much more strongly than longer red wavelengths. This scattered blue light reaches our eyes from different directions, making the sky appear blue.

Sunrise and Sunset:

At sunrise and sunset, the Sun is near the horizon, so sunlight travels a much longer distance through the atmosphere.

Much of the blue light is scattered away before reaching the observer. The longer red and orange wavelengths are scattered less and dominate the light reaching our eyes.

As a result, the sky appears reddish or orange.

11. Compare radiation pressure propulsion with chemical propulsion in space exploration. Which is more viable for long-term missions, and why?

Comparison:

Feature

Chemical Propulsion

Radiation Pressure Propulsion (Solar Sails)

Fuel Requirement

Requires onboard chemical propellant.

Uses no onboard fuel; powered by photon momentum.

Thrust Level

High thrust for short durations.

Continuous but extremely low thrust over long durations.

Mass Efficiency

Limited by the mass of the required propellant.

Has no fuel mass penalty for propulsion.

Viability for Long-Term Missions:

Radiation pressure propulsion using solar sails can be more suitable for certain long-term interplanetary and deep-space missions.

It relies on the continuous pressure exerted by sunlight photons. Since it does not require onboard propellant, it can continue accelerating over long periods without running out of fuel.

Although the thrust is very small, continuous acceleration can gradually produce significant changes in spacecraft velocity.

Numerical Response Questions

1. Calculate the wavelength of an electromagnetic wave with a frequency of \(5.0\times10^{14}\text{ Hz}\).

Given:

Frequency:

\[ f=5.0\times10^{14}\text{ Hz} \]

Speed of light:

\[ c=3.0\times10^8\text{ m/s} \]

Formula:

\[ c=f\lambda \]

Therefore:

\[ \lambda=\frac{c}{f} \]

Calculation:

\[ \lambda=\frac{3.0\times10^8}{5.0\times10^{14}} \]
\[ \lambda=6.0\times10^{-7}\text{ m} \]

Converting into nanometres:

\[ \lambda=600\text{ nm} \]

Answer:

\[ \boxed{\lambda=600\text{ nm}} \]

2. A gamma-ray photon has a frequency of \(2.0\times10^{20}\text{ Hz}\). Calculate its energy.

Given:

Frequency:

\[ f=2.0\times10^{20}\text{ Hz} \]

Planck’s constant:

\[ h=6.63\times10^{-34}\text{ J}\cdot\text{s} \]

Formula:

\[ E=hf \]

Calculation:

\[ E=(6.63\times10^{-34})(2.0\times10^{20}) \]
\[ E=1.326\times10^{-13}\text{ J} \]
\[ E\approx1.33\times10^{-13}\text{ J} \]

Answer:

\[ \boxed{E=1.33\times10^{-13}\text{ J}} \]

3. A laser emits light with a frequency of \(4.0\times10^{14}\text{ Hz}\). What is the momentum of each photon?

Given:

Frequency:

\[ f=4.0\times10^{14}\text{ Hz} \]

Planck’s constant:

\[ h=6.63\times10^{-34}\text{ J}\cdot\text{s} \]

Speed of light:

\[ c=3.0\times10^8\text{ m/s} \]

Formula:

The momentum of a photon is:

\[ p=\frac{E}{c} \]

Since:

\[ E=hf \]

Therefore:

\[ p=\frac{hf}{c} \]

Calculation:

\[ p=\frac{(6.63\times10^{-34})(4.0\times10^{14})}{3.0\times10^8} \]
\[ p=\frac{2.652\times10^{-19}}{3.0\times10^8} \]
\[ p=8.84\times10^{-28}\text{ kg}\cdot\text{m/s} \]

Answer:

\[ \boxed{p=8.84\times10^{-28}\text{ kg}\cdot\text{m/s}} \]

4. What will be the momentum and energy of a photon from radio waves with a wavelength of 100 mm?

Given:

Wavelength:

\[ \lambda=100\text{ mm}=0.1\text{ m} \]

Planck’s constant:

\[ h=6.63\times10^{-34}\text{ J}\cdot\text{s} \]

Speed of light:

\[ c=3.0\times10^8\text{ m/s} \]

Electron charge:

\[ 1\text{ eV}=1.6\times10^{-19}\text{ J} \]

Part (a): Momentum (\(p\))

Formula:

\[ p=\frac{h}{\lambda} \]

Calculation:

\[ p=\frac{6.63\times10^{-34}}{0.1} \]
\[ p=6.63\times10^{-33}\text{ kg}\cdot\text{m/s} \]

Part (b): Energy (\(E\))

Formula:

\[ E=\frac{hc}{\lambda} \]

Calculation:

\[ E=\frac{(6.63\times10^{-34})(3.0\times10^8)}{0.1} \]
\[ E=1.989\times10^{-24}\text{ J} \]

Converting joules into electron volts:

\[ E=\frac{1.989\times10^{-24}}{1.6\times10^{-19}} \]
\[ E=1.24\times10^{-5}\text{ eV} \]

Answer:

\[ \boxed{p=6.63\times10^{-33}\text{ kg}\cdot\text{m/s}} \]
\[ \boxed{E=1.24\times10^{-5}\text{ eV}} \]

5. The surface of the Earth is hit by radiation of different frequencies and wavelengths at all times, mainly during the daytime. The maximum energy of radiation received in your room during a sunny day is 4.7 eV. Find the frequency and wavelength of this radiation. Also, identify its type.

Given:

Energy:

\[ E=4.7\text{ eV} \]

Converting into joules:

\[ E=4.7\times1.6\times10^{-19} \]
\[ E=7.52\times10^{-19}\text{ J} \]

Planck’s constant:

\[ h=6.63\times10^{-34}\text{ J}\cdot\text{s} \]

Speed of light:

\[ c=3.0\times10^8\text{ m/s} \]

Part (a): Frequency (\(f\))

Formula:

\[ E=hf \]

Therefore:

\[ f=\frac{E}{h} \]

Calculation:

\[ f=\frac{7.52\times10^{-19}}{6.63\times10^{-34}} \]
\[ f=1.134\times10^{15}\text{ Hz} \]

Part (b): Wavelength (\(\lambda\))

Formula:

\[ \lambda=\frac{c}{f} \]

Calculation:

\[ \lambda=\frac{3.0\times10^8}{1.134\times10^{15}} \]
\[ \lambda=2.645\times10^{-7}\text{ m} \]

Converting into nanometres:

\[ \lambda\approx265\text{ nm} \]

Part (c): Type of Radiation

A wavelength of 265 nm lies within the ultraviolet (UV) region of the electromagnetic spectrum.

Answer:

\[ \boxed{f=1.13\times10^{15}\text{ Hz}} \]
\[ \boxed{\lambda=265\text{ nm}} \]

Type: Ultraviolet (UV) radiation.

6. Find the momentum and energy of a photon from visible light of the smallest wavelength.

Given:

Smallest wavelength of visible light (violet):

\[ \lambda=380\text{ nm}=3.8\times10^{-7}\text{ m} \]

Planck’s constant:

\[ h=6.63\times10^{-34}\text{ J}\cdot\text{s} \]

Speed of light:

\[ c=3.0\times10^8\text{ m/s} \]

Conversion factor:

\[ 1\text{ eV}=1.6\times10^{-19}\text{ J} \]

Part (a): Energy (\(E\))

Formula:

\[ E=\frac{hc}{\lambda} \]

Calculation:

\[ E=\frac{(6.63\times10^{-34})(3.0\times10^8)}{3.8\times10^{-7}} \]
\[ E=5.234\times10^{-19}\text{ J} \]

Converting into electron volts:

\[ E=\frac{5.234\times10^{-19}}{1.6\times10^{-19}} \]
\[ E\approx3.27\text{ eV} \]

Part (b): Momentum (\(p\))

Formula:

\[ p=\frac{h}{\lambda} \]

Calculation:

\[ p=\frac{6.63\times10^{-34}}{3.8\times10^{-7}} \]
\[ p=1.74\times10^{-27}\text{ kg}\cdot\text{m/s} \]

Answer:

\[ \boxed{E=3.27\text{ eV}} \]
\[ \boxed{p=1.74\times10^{-27}\text{ kg}\cdot\text{m/s}} \]
 

Notes by Sir Dilawar
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