Electromagnetic Waves and Electromagnetic Spectrum Practice Questions — JAMB UTME

Question 1Question →

An electromagnetic wave has a wavelength of $1.5 \times 10^{-7}\text{ m}$ in a vacuum. It enters an optical medium where its speed decreases to $2.0 \times 10^8\text{ m/s}$ . Given that the speed of light in vacuum is $c = 3.0 \times 10^8\text{ m/s}$ , what is the wavelength of the wave in the medium, and to which spectral region does its frequency belong?

1.0 \times 10^{-7}\text{ m}

and Ultraviolet region

1.0 \times 10^{-7}\text{ m}

and Visible light region

2.25 \times 10^{-7}\text{ m}

and Ultraviolet region

1.5 \times 10^{-7}\text{ m}

and Infrared region

Show answer & explanation

Answer: $1.0 \times 10^{-7}\text{ m}$ and Ultraviolet region

Answer

The wavelength in the medium is

1.0 \times 10^{-7}\text{ m}

and its frequency belongs to the Ultraviolet region.

When an electromagnetic wave transitions into a medium, its frequency

f

remains constant at

2.0 \times 10^{15}\text{ Hz}

, placing it strictly in the Ultraviolet spectrum. Its wavelength in the medium shrinks to

\lambda_{med} = \frac{v}{f} = \frac{2.0 \times 10^8}{2.0 \times 10^{15}} = 1.0 \times 10^{-7}\text{ m}

.

Step-by-Step Solution

1

Calculate the frequency of the wave in vacuum using the wave equation

c = f \lambda_{vac}

.

f = \frac{3.0 \times 10^8\text{ m/s}}{1.5 \times 10^{-7}\text{ m}} = 2.0 \times 10^{15}\text{ Hz}

.

Frequency is an intrinsic property determined by the source and does not change when entering a new medium.

2

Identify the electromagnetic spectrum region corresponding to the calculated frequency.

A frequency of

2.0 \times 10^{15}\text{ Hz}

falls in the range

7.5 \times 10^{14}\text{ Hz}

to

3.0 \times 10^{16}\text{ Hz}

, which corresponds to the Ultraviolet region.

Spectral classifications are uniquely determined by frequency (or vacuum wavelength).

3

Determine the wavelength in the medium using

\lambda_{med} = \frac{v}{f}

.

\lambda_{med} = \frac{2.0 \times 10^8\text{ m/s}}{2.0 \times 10^{15}\text{ Hz}} = 1.0 \times 10^{-7}\text{ m}

.

The wavelength changes proportionally with phase velocity in a medium.

Key Concept

Invariance of electromagnetic wave frequency across media boundaries and wave equation relationship

v = f \lambda

.

Estimated Time:2m 0s

Question 2Question →

A radio transmitter emits electromagnetic waves with a frequency of $1.0 \times 10^8\text{ Hz}$ . Given that the speed of light in a vacuum is $c = 3.0 \times 10^8\text{ m s}^{-1}$ , what is the wavelength of the emitted wave?

3.0\text{ m}

3.0 \times 10^{16}\text{ m}

0.33\text{ m}

3.0 \times 10^{-1}\text{ m}

Show answer & explanation

Answer: $3.0\text{ m}$

Answer

The wavelength of the emitted radio wave is

3.0\text{ m}

.

According to the wave equation

c = f \lambda

, the wavelength is found by dividing the speed of light by frequency:

\lambda = \frac{3.0 \times 10^8\text{ m s}^{-1}}{1.0 \times 10^8\text{ Hz}} = 3.0\text{ m}

.

Step-by-Step Solution

1

Identify the given parameters and fundamental formula

Speed of light

c = 3.0 \times 10^8\text{ m s}^{-1}

and frequency

f = 1.0 \times 10^8\text{ Hz}

. The electromagnetic wave relation is

c = f \lambda

.

All electromagnetic waves travel at the speed of light in a vacuum.

2

Rearrange the formula to solve for wavelength

\lambda = \frac{c}{f}

Dividing both sides of the wave equation by frequency isolates wavelength.

3

Substitute values into the equation and compute the result

\lambda = \frac{3.0 \times 10^8}{1.0 \times 10^8} = 3.0\text{ m}

Performing the division yields the correct wavelength.

Key Concept

Relationship between wave speed, frequency, and wavelength for electromagnetic radiation in a vacuum.

Question 3Question →

An infrared sensor detects electromagnetic radiation with a frequency of $4.0 \times 10^{14}\text{ Hz}$ . A second sensor detects ultraviolet radiation with a wavelength of $300\text{ nm}$ . Given that the speed of light in a vacuum is $c = 3.0 \times 10^8\text{ m/s}$ , what is the ratio of the wavelength of the infrared radiation to the wavelength of the ultraviolet radiation?

2.5

0.4

25.0

1.33

Show answer & explanation

Answer: $2.5$

Answer

The ratio of the wavelength of the infrared radiation to the wavelength of the ultraviolet radiation is 2.5.

Using

c = f \lambda

, the infrared wavelength is

\lambda = \frac{3.0 \times 10^8}{4.0 \times 10^{14}} = 7.5 \times 10^{-7}\text{ m}

. Comparing this to the ultraviolet wavelength of

300\text{ nm} = 3.0 \times 10^{-7}\text{ m}

gives a ratio of

\frac{7.5 \times 10^{-7}}{3.0 \times 10^{-7}} = 2.5

.

Step-by-Step Solution

1

Calculate the wavelength of the infrared radiation using the wave equation

c = f \lambda

.

\lambda_{\text{IR}} = \frac{3.0 \times 10^8\text{ m/s}}{4.0 \times 10^{14}\text{ Hz}} = 7.5 \times 10^{-7}\text{ m}

The speed of all electromagnetic waves in a vacuum is constant (

c = 3.0 \times 10^8\text{ m/s}

).

2

Convert the ultraviolet wavelength to meters for consistent units.

\lambda_{\text{UV}} = 300\text{ nm} = 300 \times 10^{-9}\text{ m} = 3.0 \times 10^{-7}\text{ m}

Units must be in standard meters before calculating the dimensionless ratio.

3

Compute the ratio of the infrared wavelength to the ultraviolet wavelength.

\text{Ratio} = \frac{7.5 \times 10^{-7}\text{ m}}{3.0 \times 10^{-7}\text{ m}} = 2.5

Dividing the two wavelengths in the same units yields the desired ratio.

Key Concept

Wave Equation and Electromagnetic Spectrum Properties

Estimated Time:2m 0s

Question 4Question →

A marine navigational radar system emits electromagnetic pulses with a wavelength of $3.0\text{ cm}$ . Given that the speed of electromagnetic waves in air is $3.0 \times 10^8\text{ m/s}$ , what is the frequency of the radar signal in gigahertz ( $\text{GHz}$ )?

Show answer & explanation

Answer: 10

Answer

The frequency of the radar signal is

10\text{ GHz}

.

Converting

3.0\text{ cm}

to meters gives

\lambda = 0.03\text{ m}

. Using

f = c / \lambda

,

f = (3.0 \times 10^8\text{ m/s}) / 0.03\text{ m} = 1.0 \times 10^{10}\text{ Hz}

. Dividing by

10^9\text{ Hz/GHz}

yields

10\text{ GHz}

.

Step-by-Step Solution

1

Convert the given wavelength into fundamental SI units (meters).

\lambda = 3.0\text{ cm} = 3.0 \times 10^{-2}\text{ m}

The speed of light

c

is given in meters per second, so the wavelength must be expressed in meters to keep units consistent.

2

Rearrange the electromagnetic wave speed formula

c = f \lambda

to solve for frequency

f

.

f = \frac{c}{\lambda} = \frac{3.0 \times 10^8\text{ m/s}}{3.0 \times 10^{-2}\text{ m}} = 1.0 \times 10^{10}\text{ Hz}

Frequency is the ratio of wave propagation speed to wavelength.

3

Convert the frequency from hertz to gigahertz.

f = \frac{1.0 \times 10^{10}\text{ Hz}}{10^9\text{ Hz/GHz}} = 10\text{ GHz}

The question explicitly asks for the answer in gigahertz (

\text{GHz}

).

Key Concept

Electromagnetic wave propagation equation (

c = f \lambda

) and unit metric prefixes.

Question 5Question →

A diagnostic medical device emits electromagnetic radiation with a frequency of $6.0 \times 10^{18}\text{ Hz}$ in a vacuum. Given that the speed of light in a vacuum is $c = 3.0 \times 10^8\text{ m/s}$ , what is the wavelength of this radiation, and to which part of the electromagnetic spectrum does it belong?

5.0 \times 10^{-11}\text{ m}

, X-rays

5.0 \times 10^{-11}\text{ m}

, Ultraviolet rays

2.0 \times 10^{10}\text{ m}

, X-rays

1.8 \times 10^{27}\text{ m}

, Gamma rays

Show answer & explanation

Answer: $5.0 \times 10^{-11}\text{ m}$ , X-rays

Answer

The wavelength of the radiation is

5.0 \times 10^{-11}\text{ m}

, which corresponds to X-rays.

Using the wave equation

\lambda = \frac{c}{f}

, dividing

3.0 \times 10^8\text{ m/s}

by

6.0 \times 10^{18}\text{ Hz}

yields

\lambda = 5.0 \times 10^{-11}\text{ m}

. Radiation with a wavelength of

5.0 \times 10^{-11}\text{ m}

(frequency

6.0 \times 10^{18}\text{ Hz}

) falls in the X-ray portion of the electromagnetic spectrum.

Step-by-Step Solution

1

Identify the given values and the wave equation relating speed, frequency, and wavelength.

Speed of light

c = 3.0 \times 10^8\text{ m/s}

, frequency

f = 6.0 \times 10^{18}\text{ Hz}

. Equation:

c = f \lambda \implies \lambda = \frac{c}{f}

.

Electromagnetic waves propagate in a vacuum at speed

c

.

2

Calculate the wavelength

\lambda

.

\lambda = \frac{3.0 \times 10^8\text{ m/s}}{6.0 \times 10^{18}\text{ s}^{-1}} = 0.5 \times 10^{-10}\text{ m} = 5.0 \times 10^{-11}\text{ m}.

Proper division of scientific notation terms.

3

Classify the electromagnetic spectral region based on the calculated wavelength.

Wavelengths in the range

10^{-11}\text{ m}

to

10^{-8}\text{ m}

correspond to X-rays.

X-rays typically span frequencies between

3 \times 10^{16}\text{ Hz}

and

3 \times 10^{19}\text{ Hz}

.

Key Concept

Wave equation for electromagnetic waves and spectral region identification

Question 6Question →

An optical communication sensor operates using an electromagnetic wave with a frequency of $1.5 \times 10^{14}\text{ Hz}$ in a vacuum. Given that the speed of light in vacuum is $3.0 \times 10^8\text{ m/s}$ , what is the wavelength of this electromagnetic radiation?

2.0 \times 10^{-6}\text{ m}

5.0 \times 10^5\text{ m}

4.5 \times 10^{22}\text{ m}

2.0 \times 10^{-7}\text{ m}

Show answer & explanation

Answer: $2.0 \times 10^{-6}\text{ m}$

Answer

The wavelength of the electromagnetic wave is

2.0 \times 10^{-6}\text{ m}

.

Using the electromagnetic wave relation

c = f\lambda

, dividing the speed of light (

3.0 \times 10^8\text{ m/s}

) by the given frequency (

1.5 \times 10^{14}\text{ Hz}

) correctly gives

2.0 \times 10^{-6}\text{ m}

.

Step-by-Step Solution

1

Identify given quantities and formula

Speed of light

c = 3.0 \times 10^8\text{ m/s}

, frequency

f = 1.5 \times 10^{14}\text{ Hz}

, wave equation

c = f \lambda

The fundamental wave equation relates wave speed, frequency, and wavelength for all electromagnetic waves.

2

Rearrange formula to solve for wavelength

\lambda

\lambda = \frac{c}{f}

Isolating the target unknown variable allows direct calculation.

3

Substitute values and compute

$\lambda = \frac{3.0 \times 10^8}{1.5 \times 10^{14}} = 2.0 \times 10^{-6}\text{ m}$

Dividing the coefficients (

3.0 / 1.5 = 2.0

) and subtracting the powers of ten (

8 - 14 = -6

) yields the accurate wavelength.

Key Concept

Wave Equation for Electromagnetic Waves

Question 7Question →

Arrange the following types of electromagnetic radiation in order of increasing frequency (from lowest frequency to highest frequency).

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct sequence of electromagnetic radiations in order of increasing frequency is: Microwaves, Infrared radiation, Ultraviolet radiation, and Gamma rays.

Electromagnetic waves propagate at the constant speed

c = 3.0 \times 10^8\text{ m/s}

in a vacuum. Frequency increases as wavelength decreases across the spectrum. Microwaves have the longest wavelength and lowest frequency among the choices, followed by infrared radiation, ultraviolet radiation, and finally gamma rays, which possess the shortest wavelength and highest frequency.

Step-by-Step Solution

1

Recall the arrangement of the electromagnetic spectrum in terms of frequency and wavelength.

In the electromagnetic spectrum, frequency increases in the order: Radio waves

\rightarrow

Microwaves

\rightarrow

Infrared

\rightarrow

Visible light

\rightarrow

Ultraviolet

\rightarrow

X-rays

\rightarrow

Gamma rays.

Electromagnetic wave energy

E = hf

and frequency

f = \frac{c}{\lambda}

increase as wavelength decreases.

2

Identify the relative position of each given radiation type along the frequency scale.

Microwaves (

10^9 - 10^{11}\text{ Hz}

) < Infrared (

10^{11} - 10^{14}\text{ Hz}

) < Ultraviolet (

10^{15} - 10^{16}\text{ Hz}

) < Gamma rays (

>10^{19}\text{ Hz}

).

Comparing their characteristic frequency ranges determines their exact position in the sequence.

3

Order the items from lowest to highest frequency.

1st: Microwaves, 2nd: Infrared radiation, 3rd: Ultraviolet radiation, 4th: Gamma rays.

This sequence satisfies the requirement of strictly increasing frequency.

Key Concept

Electromagnetic Spectrum Ordering by Frequency and Wavelength

Question 8Question →

An FM radio station transmits electromagnetic waves at a frequency of $1.0 \times 10^8\text{ Hz}$ in a vacuum. Given that the speed of light in a vacuum is $3.0 \times 10^8\text{ m/s}$ , what is the wavelength of these radio waves in meters?

Show answer & explanation

Answer: 3

Answer

The wavelength of the radio waves is

3.0\text{ m}

.

Applying the wave equation

c = f\lambda

, rearranging to

\lambda = \frac{c}{f}

, and substituting

c = 3.0 \times 10^8\text{ m/s}

and

f = 1.0 \times 10^8\text{ Hz}

gives a wavelength of

3.0\text{ m}

.

Step-by-Step Solution

1

Identify given parameters and key formula

Speed of light

c = 3.0 \times 10^8\text{ m/s}

, frequency

f = 1.0 \times 10^8\text{ Hz}

, wave equation

c = f\lambda

The electromagnetic wave equation relates speed, frequency, and wavelength.

2

Rearrange for wavelength and substitute given values

\lambda = \frac{c}{f} = \frac{3.0 \times 10^8}{1.0 \times 10^8} = 3.0\text{ m}

Dividing the speed of propagation by the wave frequency yields the spatial wavelength.

Key Concept

Wave equation relating speed, frequency, and wavelength of electromagnetic waves

Question 9Question →

Which type of electromagnetic radiation is primarily detected using a thermopile or a bolometer due to its heating effect?

Infrared radiation

Ultraviolet radiation

Gamma radiation

Radio waves

Show answer & explanation

Answer: Infrared radiation

Answer

Infrared radiation is the component of the electromagnetic spectrum primarily detected by a thermopile or bolometer.

Infrared radiation is absorbed by matter primarily as heat, raising the temperature of the absorber. Thermopiles measure this temperature rise by generating a small voltage across thermocouple junctions, making infrared radiation the primary band detected by thermopiles.

Step-by-Step Solution

1

Identify the characteristic mechanism of detection for thermopiles

Thermopiles convert thermal energy (heat) into an electrical voltage via the thermoelectric effect.

Understanding the working principle of the detector points to the radiation band with prominent heating properties.

2

Match the detector to the corresponding electromagnetic band

Infrared radiation causes significant thermal excitation and temperature rise upon absorption, making thermopiles ideal detectors.

Infrared rays are also known as heat waves because they transfer thermal energy efficiently.

Key Concept

Detection mechanisms of electromagnetic waves

Estimated Time:45s

Question 10Question →

An electromagnetic wave propagating in a vacuum has a frequency of $6.0 \times 10^{14}\text{ Hz}$ . It passes from the vacuum into a dense glass block with a refractive index of $1.50$ . Given that the speed of light in vacuum is $c = 3.0 \times 10^8\text{ m/s}$ , what is the wavelength of the wave inside the glass block, and to which region of the electromagnetic spectrum does the wave belong based on its vacuum properties?

3.33 \times 10^{-7}\text{ m}

, Visible light

5.00 \times 10^{-7}\text{ m}

, Visible light

7.50 \times 10^{-7}\text{ m}

, Infrared

2.25 \times 10^{-7}\text{ m}

, Ultraviolet

Show answer & explanation

Answer: $3.33 \times 10^{-7}\text{ m}$ , Visible light

Answer

The wavelength of the wave inside the glass block is

3.33 \times 10^{-7}\text{ m}

, and the wave belongs to the Visible light region.

The vacuum wavelength of the wave is

\lambda_0 = c/f = 5.0 \times 10^{-7}\text{ m}

, placing it in the visible light spectrum. Upon entering the glass medium (

n = 1.50

), the wave frequency remains unchanged while its wavelength is compressed by the factor

n

, giving

\lambda = (5.0 \times 10^{-7})/1.50 = 3.33 \times 10^{-7}\text{ m}

.

Step-by-Step Solution

1

Calculate the vacuum wavelength of the electromagnetic wave

\lambda_0 = \frac{c}{f} = \frac{3.0 \times 10^8\text{ m/s}}{6.0 \times 10^{14}\text{ Hz}} = 5.0 \times 10^{-7}\text{ m}

The wave equation in vacuum relates wave speed, frequency, and wavelength by

c = f \lambda_0

.

2

Classify the electromagnetic spectral region

Visible light region

A vacuum wavelength of

5.0 \times 10^{-7}\text{ m}

(

500\text{ nm}

) falls within the visible light band (

400\text{ nm} - 700\text{ nm}

).

3

Calculate the wavelength inside the glass block

\lambda = \frac{\lambda_0}{n} = \frac{5.0 \times 10^{-7}\text{ m}}{1.50} = 3.33 \times 10^{-7}\text{ m}

When passing into a medium with refractive index

n

, frequency remains constant while the wave speed and wavelength are reduced by a factor of

n

.

Key Concept

Electromagnetic Wave Refraction and Spectrum Classification

Question 11Question →

Arrange the following regions of the electromagnetic spectrum in order of decreasing wavelength (from longest wavelength to shortest wavelength).

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct sequence from longest to shortest wavelength is: Microwaves, Infrared radiation, Ultraviolet radiation, and Gamma rays.

The electromagnetic spectrum ordered by decreasing wavelength (longest to shortest) follows the sequence: radio waves/microwaves, infrared, visible light, ultraviolet, X-rays, and gamma rays. Thus, microwaves come first with the longest wavelength, followed by infrared, ultraviolet, and finally gamma rays with the shortest wavelength.

Step-by-Step Solution

1

Recall the wave equation

c = f \lambda

connecting frequency (

f

) and wavelength (

\lambda

) for electromagnetic waves in a vacuum.

Wavelength is inversely proportional to frequency and photon energy.

Since the speed of light

c

is constant, waves with lower frequencies have longer wavelengths.

2

Identify the relative wavelengths of each specified region of the electromagnetic spectrum.

Microwaves (

10^{-3}\text{ m} - 10^{-1}\text{ m}

) > Infrared (

7 \times 10^{-7}\text{ m} - 10^{-3}\text{ m}

) > Ultraviolet (

10^{-8}\text{ m} - 4 \times 10^{-7}\text{ m}

) > Gamma rays (

< 10^{-11}\text{ m}

).

Microwaves sit near the radio end of the spectrum, while gamma rays lie at the extreme high-energy end.

3

Sequence the items from longest wavelength to shortest wavelength.

Microwaves

\rightarrow

Infrared radiation

\rightarrow

Ultraviolet radiation

\rightarrow

Gamma rays.

This arranges the waves in strict order of decreasing wavelength.

Key Concept

Electromagnetic spectrum wavelength and frequency hierarchy

Question 12Question →

An electromagnetic microwave signal used in telecommunication has a wavelength of $0.02\text{ m}$ in a vacuum. Given that the speed of light in a vacuum is $3.0 \times 10^8\text{ m/s}$ , calculate the frequency of the signal in gigahertz ( $\text{GHz}$ ).

Show answer & explanation

Answer: 15

Answer

The frequency of the microwave signal is

15\text{ GHz}

.

Using the electromagnetic wave equation

c = f \lambda

, the frequency in Hz is calculated as

f = \frac{c}{\lambda} = \frac{3.0 \times 10^8\text{ m/s}}{0.02\text{ m}} = 1.5 \times 10^{10}\text{ Hz}

. Dividing by

10^9

to convert into gigahertz gives

15\text{ GHz}

.

Step-by-Step Solution

1

Identify the given parameters and formula.

Speed of light

c = 3.0 \times 10^8\text{ m/s}

, wavelength

\lambda = 0.02\text{ m}

, and wave equation

c = f \lambda

.

Electromagnetic waves propagate at speed

c

in a vacuum, relating frequency and wavelength.

2

Calculate the frequency in Hertz (Hz).

f = \frac{3.0 \times 10^8}{0.02} = 1.5 \times 10^{10}\text{ Hz}

.

Dividing the wave speed by the wavelength yields the frequency.

3

Convert the unit from Hz to GHz.

\frac{1.5 \times 10^{10}\text{ Hz}}{10^9\text{ Hz/GHz}} = 15\text{ GHz}

.

One gigahertz (

1\text{ GHz}

) equals

10^9\text{ Hz}

.

Key Concept

Relationship between speed of light, frequency, and wavelength (

c = f \lambda

) for electromagnetic radiation.

Question 13Question →

The electromagnetic spectrum consists of waves with varying frequencies, wavelengths, and photon energies, each associated with distinct physical detection mechanisms and applications. Consider the following types of electromagnetic radiation:

I. Radiation emitted by warm bodies, primarily detected using a thermopile.
II. Radiation utilized in radar systems and satellite communications.
III. Radiation emitted during nuclear decay processes, detected using a Geiger-Müller counter.
IV. Radiation responsible for sun tanning and detected by its ability to induce fluorescence on zinc sulfide screens.

Arrange these four types of electromagnetic radiation in order of increasing photon energy (from lowest photon energy to highest photon energy).

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct sequence in order of increasing photon energy is: Radiation utilized in radar systems (Microwaves) < Radiation emitted by warm bodies (Infrared) < Radiation causing sun tanning (Ultraviolet) < Radiation emitted during nuclear decay (Gamma rays).

Microwaves possess the lowest frequency among the four types, followed by infrared radiation, then ultraviolet radiation, and finally gamma rays which possess the highest frequency and photon energy.

Step-by-Step Solution

1

Identify the region of the electromagnetic spectrum corresponding to each property and detector described.

Item I corresponds to Infrared radiation; Item II corresponds to Microwaves; Item III corresponds to Gamma rays; Item IV corresponds to Ultraviolet radiation.

Thermopiles detect thermal radiation (IR); radar uses microwaves; Geiger-Müller counters detect nuclear ionizing radiation (Gamma rays); fluorescence on ZnS is caused by UV light.

2

Relate photon energy

E

to frequency

f

and wavelength

\lambda

using Planck's relation

E = hf = \frac{hc}{\lambda}

.

Photon energy is directly proportional to frequency (

E \propto f

) and inversely proportional to wavelength (

E \propto \frac{1}{\lambda}

).

Higher frequency radiation consists of more energetic individual photons.

3

Sequence the identified electromagnetic waves from lowest frequency to highest frequency.

Microwaves (

f \approx 10^9 - 10^{11}\text{ Hz}

) < Infrared (

f \approx 10^{11} - 4 \times 10^{14}\text{ Hz}

) < Ultraviolet (

f \approx 7.5 \times 10^{14} - 3 \times 10^{16}\text{ Hz}

) < Gamma rays (

f > 10^{19}\text{ Hz}

).

This sequence reflects the fundamental order of increasing photon energy across the spectrum.

Key Concept

Electromagnetic Spectrum Spectral Regions, Detection Devices, and Photon Energy Ordering

Question 14Question →

A physics laboratory utilizes four specialized instruments to detect different regions of the electromagnetic spectrum: an aerial antenna, a thermopile, a photographic plate sensitive to sun-tanning radiation, and a Geiger-Müller tube.

Arrange these detectors in order of INCREASING frequency of the electromagnetic radiation they are primarily designed to detect (from lowest frequency to highest frequency).

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct order from lowest frequency to highest frequency is: Aerial antenna (Radio waves) → Thermopile (Infrared) → Photographic plate sensitive to sun-tanning radiation (Ultraviolet) → Geiger-Müller tube (Gamma rays).

The correct sequence arranges the instruments according to the increasing frequency of the radiation they detect. Radio waves (detected by an aerial antenna) have the lowest frequency, followed by infrared radiation (detected by a thermopile), ultraviolet radiation (detected by photographic plates sensitive to sun-tanning rays), and gamma rays (detected by a Geiger-Müller tube) which have the highest frequency.

Step-by-Step Solution

1

Identify the type of electromagnetic radiation detected by each device.

Aerial antenna detects radio waves; Thermopile detects infrared radiation; Photographic plate for tanning radiation detects ultraviolet radiation; Geiger-Müller tube detects gamma rays.

Each detector operates on specific physical properties characteristic of a particular band of the electromagnetic spectrum.

2

Recall the order of the electromagnetic spectrum in terms of frequency (

f

).

Radio waves (

< 10^9\text{ Hz}

) < Infrared (

10^{11} - 10^{14}\text{ Hz}

) < Ultraviolet (

10^{15} - 10^{17}\text{ Hz}

) < Gamma rays (

> 10^{19}\text{ Hz}

).

Frequency increases continuously across the spectrum from radio waves to gamma rays.

3

Sequence the detectors based on their associated radiation frequencies from lowest to highest.

Aerial antenna

\rightarrow

Thermopile

\rightarrow

Photographic plate sensitive to sun-tanning radiation

\rightarrow

Geiger-Müller tube.

This directly matches the increasing frequency order of radio waves, infrared, ultraviolet, and gamma rays.

Key Concept

Detection mechanisms and frequency distribution across the electromagnetic spectrum

Estimated Time:2m 0s

Question 15Question →

Arrange the following electromagnetic radiation applications in order of increasing photon energy, starting from the radiation with the lowest energy to the one with the highest energy.

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct sequence from lowest to highest photon energy is: Radar waves (microwaves), radiant heat (infrared), sterilizing radiation (ultraviolet), and nuclear gamma emissions.

The photon energy of electromagnetic radiation is directly proportional to its frequency (

E = hf

). Microwaves have the lowest frequency among the listed types, followed by infrared radiation, then ultraviolet radiation, with gamma rays having the highest frequency and thus the greatest photon energy per photon.

Step-by-Step Solution

1

Identify the region of the electromagnetic spectrum for each listed application.

Radar waves belong to microwaves; radiant heat corresponds to infrared radiation; sterilization uses ultraviolet light; nuclear emissions are gamma rays.

Connecting applications to their respective spectral regions is required to compare physical properties.

2

Recall the relationship between frequency and photon energy in the electromagnetic spectrum using

E = h f

.

Photon energy is directly proportional to frequency (

E \propto f

), meaning higher frequency waves carry greater energy per photon.

Understanding quantum photon energy helps determine the correct energy ranking.

3

Order the spectral regions from lowest frequency to highest frequency.

The order of increasing frequency (and energy) is: Microwaves < Infrared < Ultraviolet < Gamma rays.

This matches the physical ordering of the electromagnetic spectrum by increasing frequency.

Key Concept

Photon energy across the electromagnetic spectrum increases with increasing frequency (

E = h f

).

Question 16Question →

A satellite communication system transmits an ultra-high frequency electromagnetic wave with a wavelength of $0.05\text{ m}$ . If the wave travels at a speed of $3.0 \times 10^{8}\text{ m/s}$ in a vacuum, what is the frequency of the transmission in gigahertz ( $\text{GHz}$ )?

Show answer & explanation

Answer: 6

Answer

The frequency of the electromagnetic transmission is

6\text{ GHz}

.

Using the wave equation

c = f \lambda

, the frequency is

f = \frac{c}{\lambda} = \frac{3.0 \times 10^8\text{ m/s}}{0.05\text{ m}} = 6.0 \times 10^9\text{ Hz}

. Converting to gigahertz (

1\text{ GHz} = 10^9\text{ Hz}

) yields

6\text{ GHz}

.

Step-by-Step Solution

1

Identify the relationship between electromagnetic wave velocity, frequency, and wavelength.

c = f \lambda

, where

c = 3.0 \times 10^8\text{ m/s}

and

\lambda = 0.05\text{ m}

.

All electromagnetic waves travel at the speed of light

c

in a vacuum.

2

Rearrange the wave equation to isolate frequency (

f

).

f = \frac{c}{\lambda} = \frac{3.0 \times 10^8\text{ m/s}}{0.05\text{ m}} = 6.0 \times 10^9\text{ Hz}

.

Dividing the wave speed by the wavelength yields the frequency in hertz.

3

Convert the frequency into gigahertz (

\text{GHz}

).

\frac{6.0 \times 10^9\text{ Hz}}{10^9\text{ Hz/GHz}} = 6\text{ GHz}

.

The prefix giga (G) denotes a factor of

10^9

.

Key Concept

Electromagnetic wave propagation equation (

c = f \lambda

) and unit conversion

Estimated Time:1m 15s

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