Problem 1
EASY
Calculate the wavelength of a radio wave with frequency 100 MHz.
Given:
\(f = 100\) MHz = \(100 \times 10^6\) Hz = \(10^8\) Hz
\(c = 3 \times 10^8\) m/s
Find: \(\lambda = ?\)
\(f = 100\) MHz = \(100 \times 10^6\) Hz = \(10^8\) Hz
\(c = 3 \times 10^8\) m/s
Find: \(\lambda = ?\)
Step-by-Step Solution:
Step 1: Write the wave equation
\[c = \lambda f \Rightarrow \lambda = \frac{c}{f}\]
Step 2: Substitute the values
\[\lambda = \frac{3 \times 10^8}{10^8}\]
Step 3: Calculate
\[\lambda = 3 \text{ m}\]
\[\boxed{\lambda = 3 \text{ m}}\]
Radio Wave – Used in FM radio broadcasting
Problem 2
EASY
Find the frequency of visible light with wavelength 600 nm (orange light).
Given:
\(\lambda = 600\) nm = \(600 \times 10^{-9}\) m = \(6 \times 10^{-7}\) m
\(c = 3 \times 10^8\) m/s
Find: \(f = ?\)
\(\lambda = 600\) nm = \(600 \times 10^{-9}\) m = \(6 \times 10^{-7}\) m
\(c = 3 \times 10^8\) m/s
Find: \(f = ?\)
Step 1: Use wave equation
\[f = \frac{c}{\lambda}\]
Step 2: Substitute values
\[f = \frac{3 \times 10^8}{6 \times 10^{-7}}\]
Step 3: Calculate
\[f = \frac{3 \times 10^8}{6 \times 10^{-7}} = 0.5 \times 10^{15} = 5 \times 10^{14} \text{ Hz}\]
\[\boxed{f = 5 \times 10^{14} \text{ Hz = 500 THz}}\]
Visible Light – Orange color in spectrum
Problem 3
MEDIUM
Calculate the time period of an electromagnetic wave with frequency 2.4 GHz (used in WiFi).
Given:
\(f = 2.4\) GHz = \(2.4 \times 10^9\) Hz
Find: \(T = ?\)
\(f = 2.4\) GHz = \(2.4 \times 10^9\) Hz
Find: \(T = ?\)
Step 1: Use the relationship between period and frequency
\[T = \frac{1}{f}\]
Step 2: Substitute the frequency
\[T = \frac{1}{2.4 \times 10^9}\]
Step 3: Calculate
\[T = \frac{1}{2.4 \times 10^9} = 0.417 \times 10^{-9} = 4.17 \times 10^{-10} \text{ s}\]
\[\boxed{T = 4.17 \times 10^{-10} \text{ s = 0.417 ns}}\]
Microwave – WiFi, Bluetooth frequency
Problem 4
EASY
An AM radio station broadcasts at 1200 kHz. Find the wavelength of these radio waves.
Given:
\(f = 1200\) kHz = \(1.2 \times 10^6\) Hz
\(c = 3 \times 10^8\) m/s
\(f = 1200\) kHz = \(1.2 \times 10^6\) Hz
\(c = 3 \times 10^8\) m/s
Solution:
\[\lambda = \frac{c}{f} = \frac{3 \times 10^8}{1.2 \times 10^6} = 250 \text{ m}\]
\[\boxed{\lambda = 250 \text{ m}}\]
AM Radio – Medium wave band
Problem 5
MEDIUM
Calculate the energy of a photon with wavelength 400 nm (violet light). Use \(h = 6.63 \times 10^{-34}\) Jā
s.
Given:
\(\lambda = 400\) nm = \(4 \times 10^{-7}\) m
\(h = 6.63 \times 10^{-34}\) Jā s
\(c = 3 \times 10^8\) m/s
\(\lambda = 400\) nm = \(4 \times 10^{-7}\) m
\(h = 6.63 \times 10^{-34}\) Jā s
\(c = 3 \times 10^8\) m/s
Step 1: Find frequency first
\[f = \frac{c}{\lambda} = \frac{3 \times 10^8}{4 \times 10^{-7}} = 7.5 \times 10^{14} \text{ Hz}\]
Step 2: Calculate energy using E = hf
\[E = hf = 6.63 \times 10^{-34} \times 7.5 \times 10^{14}\]
Step 3: Calculate final value
\[E = 49.73 \times 10^{-20} = 4.97 \times 10^{-19} \text{ J}\]
\[\boxed{E = 4.97 \times 10^{-19} \text{ J = 3.11 eV}}\]
Violet Light – Highest energy visible light
Problem 6
HARD
A mobile phone operates at 900 MHz. If the phone transmits for 3 complete wave cycles, find the total time duration and the total distance traveled by the wave.
Given:
\(f = 900\) MHz = \(9 \times 10^8\) Hz
Number of cycles = 3
Find: Time duration and distance
\(f = 900\) MHz = \(9 \times 10^8\) Hz
Number of cycles = 3
Find: Time duration and distance
Step 1: Find time period of one cycle
\[T = \frac{1}{f} = \frac{1}{9 \times 10^8} = 1.11 \times 10^{-9} \text{ s}\]
Step 2: Calculate total time for 3 cycles
\[\text{Total time} = 3T = 3 \times 1.11 \times 10^{-9} = 3.33 \times 10^{-9} \text{ s}\]
Step 3: Find wavelength
\[\lambda = \frac{c}{f} = \frac{3 \times 10^8}{9 \times 10^8} = 0.333 \text{ m}\]
Step 4: Calculate total distance for 3 wavelengths
\[\text{Total distance} = 3\lambda = 3 \times 0.333 = 1 \text{ m}\]
\[\boxed{\text{Time} = 3.33 \text{ ns}, \text{ Distance} = 1 \text{ m}}\]
Mobile Phone – GSM frequency band
Problem 7
EASY
Find the frequency of red light with wavelength 700 nm.
Given:
\(\lambda = 700\) nm = \(7 \times 10^{-7}\) m
\(\lambda = 700\) nm = \(7 \times 10^{-7}\) m
Solution:
\[f = \frac{c}{\lambda} = \frac{3 \times 10^8}{7 \times 10^{-7}} = 4.29 \times 10^{14} \text{ Hz}\]
\[\boxed{f = 4.29 \times 10^{14} \text{ Hz = 429 THz}}\]
Red Light – Longest wavelength visible light
Problem 8
MEDIUM
An X-ray has a frequency of \(3 \times 10^{18}\) Hz. Calculate its wavelength and identify its type in the EM spectrum.
Calculate wavelength:
\[\lambda = \frac{c}{f} = \frac{3 \times 10^8}{3 \times 10^{18}} = 10^{-10} \text{ m} = 0.1 \text{ nm}\]
Identification:
Wavelength of 0.1 nm falls in the X-ray region (0.01 – 10 nm)
\[\boxed{\lambda = 0.1 \text{ nm}}\]
X-Ray – Used in medical imaging
Problem 9
HARD
Compare the energies of radio waves (f = 100 MHz) and gamma rays (f = \(10^{20}\) Hz). Find the ratio of their energies.
Step 1: Energy formula
\[E = hf\]
Since h is constant, \(E \propto f\)
Step 2: Calculate energy ratio
\[\frac{E_{\gamma}}{E_{radio}} = \frac{f_{\gamma}}{f_{radio}} = \frac{10^{20}}{10^8} = 10^{12}\]
\[\boxed{\text{Gamma rays have } 10^{12} \text{ times more energy than radio waves}}\]
This is a trillion times difference in energy!
Problem 10
MEDIUM
A microwave oven operates at 2450 MHz. Calculate the wavelength and time period of these microwaves.
Given: \(f = 2450\) MHz = \(2.45 \times 10^9\) Hz
Wavelength:
\[\lambda = \frac{3 \times 10^8}{2.45 \times 10^9} = 0.122 \text{ m} = 12.2 \text{ cm}\]
Time period:
\[T = \frac{1}{f} = \frac{1}{2.45 \times 10^9} = 4.08 \times 10^{-10} \text{ s}\]
\[\boxed{\lambda = 12.2 \text{ cm}, \, T = 0.408 \text{ ns}}\]
Microwave – Perfect for heating water molecules
Problem 11
EASY
Calculate the frequency of infrared radiation with wavelength 10 Ī¼m.
\[f = \frac{3 \times 10^8}{10 \times 10^{-6}} = 3 \times 10^{13} \text{ Hz = 30 THz}\]
Infrared – Thermal imaging range
Problem 12
MEDIUM
UV light has frequency \(7.5 \times 10^{14}\) Hz. Find its wavelength and energy per photon.
\[\lambda = \frac{3 \times 10^8}{7.5 \times 10^{14}} = 4 \times 10^{-7} \text{ m = 400 nm}\]
\[E = 6.63 \times 10^{-34} \times 7.5 \times 10^{14} = 4.97 \times 10^{-19} \text{ J}\]
\[\boxed{\lambda = 400 \text{ nm}, \, E = 4.97 \times 10^{-19} \text{ J}}\]
UV Light – UV-A range
Problem 13
EASY
Find the time period of yellow light with frequency \(5.2 \times 10^{14}\) Hz.
\[T = \frac{1}{5.2 \times 10^{14}} = 1.92 \times 10^{-15} \text{ s = 1.92 fs}\]
Yellow Light – Human eye most sensitive
Problem 14
HARD
A satellite communication uses 12 GHz frequency. How many complete wavelengths fit in a 100 m transmission path?
\[\lambda = \frac{3 \times 10^8}{12 \times 10^9} = 0.025 \text{ m = 2.5 cm}\]
\[\text{Number of wavelengths} = \frac{100}{0.025} = 4000\]
\[\boxed{4000 \text{ complete wavelengths}}\]
Satellite Communication – Ku-band frequency
Problem 15
MEDIUM
Green light (Ī» = 550 nm) and blue light (Ī» = 450 nm). Find the ratio of their frequencies.
Since \(f = \frac{c}{\lambda}\) and c is constant:
\[\frac{f_{blue}}{f_{green}} = \frac{\lambda_{green}}{\lambda_{blue}} = \frac{550}{450} = 1.22\]
\[\boxed{f_{blue} : f_{green} = 1.22 : 1}\]
Blue light has 22% higher frequency than green
Problem 16
EASY
FM radio operates at 101.5 MHz. Calculate the wavelength.
\[\lambda = \frac{3 \times 10^8}{101.5 \times 10^6} = 2.96 \text{ m}\]
FM Radio – VHF band
Problem 17
HARD
Calculate the energy difference between red light (700 nm) and violet light (400 nm) photons.
\[E_{red} = \frac{hc}{\lambda_{red}} = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{7 \times 10^{-7}} = 2.84 \times 10^{-19} \text{ J}\]
\[E_{violet} = \frac{hc}{\lambda_{violet}} = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{4 \times 10^{-7}} = 4.97 \times 10^{-19} \text{ J}\]
\[\Delta E = 4.97 \times 10^{-19} – 2.84 \times 10^{-19} = 2.13 \times 10^{-19} \text{ J}\]
\[\boxed{\Delta E = 2.13 \times 10^{-19} \text{ J = 1.33 eV}}\]
Problem 18
MEDIUM
A radar operates at 10 GHz. Find the time for one complete oscillation.
\[T = \frac{1}{10 \times 10^9} = 10^{-10} \text{ s = 0.1 ns}\]
Radar – X-band frequency
Problem 19
EASY
Calculate the wavelength of gamma rays with frequency \(10^{20}\) Hz.
\[\lambda = \frac{3 \times 10^8}{10^{20}} = 3 \times 10^{-12} \text{ m = 3 pm}\]
Gamma Ray – Nuclear medicine range
Problem 20
HARD
A TV broadcast uses 200 MHz frequency. If the TV transmits continuously for 1 hour, how many complete wave cycles are transmitted?
Given: f = 200 MHz = \(2 \times 10^8\) Hz, t = 1 hour = 3600 s
\[\text{Number of cycles} = f \times t = 2 \times 10^8 \times 3600\]
\[= 7.2 \times 10^{11} \text{ cycles}\]
\[\boxed{7.2 \times 10^{11} \text{ complete cycles}}\]
That’s 720 billion wave cycles in one hour!
ā” Exam Tips & Key Relationships
š¢ Essential Formulas:
\(c = \lambda f\)
\(T = \frac{1}{f}\)
\(E = hf = \frac{hc}{\lambda}\)
\(c = \lambda f\)
\(T = \frac{1}{f}\)
\(E = hf = \frac{hc}{\lambda}\)
š Unit Conversions:
1 nm = \(10^{-9}\) m
1 Ī¼m = \(10^{-6}\) m
1 MHz = \(10^6\) Hz
1 GHz = \(10^9\) Hz
1 nm = \(10^{-9}\) m
1 Ī¼m = \(10^{-6}\) m
1 MHz = \(10^6\) Hz
1 GHz = \(10^9\) Hz
š Spectrum Memory:
Royal Men In Very Ugly X-mas Gowns
(Radio, Microwave, IR, Visible, UV, X-ray, Gamma)
Royal Men In Very Ugly X-mas Gowns
(Radio, Microwave, IR, Visible, UV, X-ray, Gamma)
ā” Speed of Light:
c = 3 Ć 10āø m/s
Remember: same for all EM waves in vacuum
c = 3 Ć 10āø m/s
Remember: same for all EM waves in vacuum
š Inverse Relationships:
ā Frequency āŗ ā Wavelength
ā Energy āŗ ā Frequency
ā Frequency āŗ ā Time Period
ā Frequency āŗ ā Wavelength
ā Energy āŗ ā Frequency
ā Frequency āŗ ā Time Period
š” Quick Calculations:
For visible light: Ī» ā 400-700 nm
For radio: Ī» > 1 m
For X-rays: Ī» < 10 nm
For visible light: Ī» ā 400-700 nm
For radio: Ī» > 1 m
For X-rays: Ī» < 10 nm