The Radio Waves Formula describes the fundamental relationships governing electromagnetic waves in the radio frequency range, expressed as \ ( c = f\lambda \), where c is the speed of light, f is frequency, and \lambda is wavelength. Radio waves are a critical topic in NCERT Class 12 Physics, Chapter 8 (Electromagnetic Waves), and they appear regularly in CBSE board exams as well as JEE Main and NEET. This article covers every important formula, derivation, solved examples at three difficulty levels, exam tips, common mistakes, and competitive exam applications — all in one place.
Key Radio Waves Formulas at a Glance
Quick reference for the most important radio waves formulas used in Class 12 and competitive exams.
- Speed-frequency-wavelength: \( c = f\lambda \)
- Frequency from wavelength: \( f = \dfrac{c}{\lambda} \)
- Wavelength from frequency: \( \lambda = \dfrac{c}{f} \)
- Energy of a photon: \( E = hf = \dfrac{hc}{\lambda} \)
- Wave number: \( k = \dfrac{2\pi}{\lambda} \)
- Angular frequency: \( \omega = 2\pi f \)
- Time period: \( T = \dfrac{1}{f} \)
What is Radio Waves Formula?
The Radio Waves Formula refers to the set of mathematical expressions that describe the properties and behaviour of radio waves — the lowest-frequency, longest-wavelength members of the electromagnetic spectrum. Radio waves occupy frequencies from about 3 Hz to 300 GHz, corresponding to wavelengths from 1 mm to 100,000 km.
In NCERT Class 12 Physics, Chapter 8 (Electromagnetic Waves), students learn that all electromagnetic waves — including radio waves — travel through a vacuum at the speed of light, \( c = 3 \times 10^8 \) m/s. The core Radio Waves Formula connects three measurable quantities: the speed of propagation, the frequency, and the wavelength.
Radio waves are produced by accelerating electric charges, oscillating LC circuits, and antennas. They are used in AM and FM broadcasting, television transmission, satellite communication, radar, and mobile networks. Understanding the formula is essential not only for CBSE board exams but also for solving problems in JEE Main and NEET, where electromagnetic spectrum questions are common.
James Clerk Maxwell first predicted electromagnetic waves theoretically in 1865, and Heinrich Hertz experimentally demonstrated radio waves in 1887. Their work forms the theoretical backbone of this topic in the NCERT curriculum.
Radio Waves Formula — Expression and Variables
The primary Radio Waves Formula is the wave equation for electromagnetic radiation:
\[ c = f \lambda \]
This can be rearranged to find frequency or wavelength separately:
\[ f = \frac{c}{\lambda} \qquad \text{and} \qquad \lambda = \frac{c}{f} \]
The energy carried by each photon of a radio wave is given by Planck’s relation:
\[ E = hf = \frac{hc}{\lambda} \]
The angular frequency and wave number are defined as:
\[ \omega = 2\pi f \qquad \text{and} \qquad k = \frac{2\pi}{\lambda} \]
Using these, the wave speed can also be written as \( c = \dfrac{\omega}{k} \).
| Symbol | Quantity | SI Unit |
|---|---|---|
| \( c \) | Speed of light (speed of radio wave in vacuum) | m/s (\( 3 \times 10^8 \) m/s) |
| \( f \) | Frequency of the radio wave | Hertz (Hz) |
| \( \lambda \) | Wavelength of the radio wave | Metre (m) |
| \( E \) | Energy of a photon | Joule (J) |
| \( h \) | Planck’s constant (\( 6.626 \times 10^{-34} \) J·s) | J·s |
| \( \omega \) | Angular frequency | rad/s |
| \( k \) | Wave number | rad/m |
| \( T \) | Time period | Second (s) |
Derivation of the Radio Waves Formula
Maxwell’s equations predict that electromagnetic waves travel at a speed given by:
\[ c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}} \]
Here, \( \mu_0 = 4\pi \times 10^{-7} \) H/m is the permeability of free space and \( \varepsilon_0 = 8.854 \times 10^{-12} \) F/m is the permittivity of free space. Substituting these values gives \( c \approx 3 \times 10^8 \) m/s.
For any wave, the relationship between speed, frequency, and wavelength comes from the definition of wave speed: in one time period \( T \), the wave travels exactly one wavelength \( \lambda \). Therefore, \( c = \lambda / T = \lambda f \), which gives the familiar Radio Waves Formula \( c = f\lambda \).
Complete Electromagnetic Waves Formula Sheet
| Formula Name | Expression | Variables | SI Units | NCERT Chapter |
|---|---|---|---|---|
| Wave Equation (Radio Waves) | \( c = f\lambda \) | c = speed of light, f = frequency, \(\lambda\) = wavelength | m/s, Hz, m | Class 12, Ch 8 |
| Photon Energy | \( E = hf \) | h = Planck’s constant, f = frequency | Joule (J) | Class 12, Ch 8 |
| Photon Energy (wavelength form) | \( E = \dfrac{hc}{\lambda} \) | h = Planck’s constant, c = speed of light, \(\lambda\) = wavelength | Joule (J) | Class 12, Ch 8 |
| Angular Frequency | \( \omega = 2\pi f \) | f = frequency | rad/s | Class 12, Ch 8 |
| Wave Number | \( k = \dfrac{2\pi}{\lambda} \) | \(\lambda\) = wavelength | rad/m | Class 12, Ch 8 |
| Wave Speed (angular form) | \( c = \dfrac{\omega}{k} \) | \(\omega\) = angular frequency, k = wave number | m/s | Class 12, Ch 8 |
| Time Period | \( T = \dfrac{1}{f} \) | f = frequency | Second (s) | Class 12, Ch 8 |
| Speed from Permittivity and Permeability | \( c = \dfrac{1}{\sqrt{\mu_0 \varepsilon_0}} \) | \(\mu_0\) = permeability, \(\varepsilon_0\) = permittivity of free space | m/s | Class 12, Ch 8 |
| Intensity of EM Wave | \( I = \dfrac{1}{2} c \varepsilon_0 E_0^2 \) | \(E_0\) = peak electric field, \(\varepsilon_0\) = permittivity | W/m² | Class 12, Ch 8 |
| Relationship between E and B fields | \( E_0 = c B_0 \) | \(E_0\) = peak electric field, \(B_0\) = peak magnetic field | V/m, T | Class 12, Ch 8 |
Radio Waves Formula — Solved Examples
Example 1 (Class 9-10 Level): Finding Wavelength from Frequency
Problem: An FM radio station broadcasts at a frequency of 100 MHz. Calculate the wavelength of the radio waves it emits.
Given:
- Frequency, \( f = 100 \text{ MHz} = 100 \times 10^6 \text{ Hz} = 10^8 \text{ Hz} \)
- Speed of light, \( c = 3 \times 10^8 \text{ m/s} \)
Step 1: Write the Radio Waves Formula: \( c = f\lambda \)
Step 2: Rearrange for wavelength: \( \lambda = \dfrac{c}{f} \)
Step 3: Substitute values: \( \lambda = \dfrac{3 \times 10^8}{10^8} = 3 \text{ m} \)
Answer
The wavelength of the FM radio wave is 3 metres.
Example 2 (Class 11-12 Level): Finding Frequency and Energy of a Radio Photon
Problem: An AM radio station transmits at a wavelength of 300 m. Find (a) the frequency of the wave and (b) the energy of one photon of this radio wave. (Use \( h = 6.626 \times 10^{-34} \) J·s, \( c = 3 \times 10^8 \) m/s.)
Given:
- Wavelength, \( \lambda = 300 \text{ m} \)
- \( c = 3 \times 10^8 \text{ m/s} \)
- \( h = 6.626 \times 10^{-34} \text{ J·s} \)
Step 1: Find frequency using \( f = \dfrac{c}{\lambda} \)
\( f = \dfrac{3 \times 10^8}{300} = 1 \times 10^6 \text{ Hz} = 1 \text{ MHz} \)
Step 2: Find photon energy using \( E = hf \)
\( E = 6.626 \times 10^{-34} \times 1 \times 10^6 \)
\( E = 6.626 \times 10^{-28} \text{ J} \)
Step 3: Verify using the alternate formula \( E = \dfrac{hc}{\lambda} \)
\( E = \dfrac{6.626 \times 10^{-34} \times 3 \times 10^8}{300} = 6.626 \times 10^{-28} \text{ J} \) ✓
Answer
(a) Frequency = 1 MHz (b) Energy per photon = \( 6.626 \times 10^{-28} \) J
Example 3 (JEE/NEET Level): Wave Number, Angular Frequency, and E-B Field Relationship
Problem: A radio wave has a frequency of 90 MHz. Calculate (a) its angular frequency \(\omega\), (b) its wave number \(k\), and (c) the peak magnetic field \(B_0\) if the peak electric field \(E_0 = 0.05 \) V/m. (Use \( c = 3 \times 10^8 \) m/s.)
Given:
- \( f = 90 \text{ MHz} = 9 \times 10^7 \text{ Hz} \)
- \( E_0 = 0.05 \text{ V/m} \)
- \( c = 3 \times 10^8 \text{ m/s} \)
Step 1: Calculate angular frequency.
\( \omega = 2\pi f = 2 \times 3.14159 \times 9 \times 10^7 \)
\( \omega \approx 5.655 \times 10^8 \text{ rad/s} \)
Step 2: Calculate wavelength first, then wave number.
\( \lambda = \dfrac{c}{f} = \dfrac{3 \times 10^8}{9 \times 10^7} = \dfrac{10}{3} \approx 3.33 \text{ m} \)
\( k = \dfrac{2\pi}{\lambda} = \dfrac{2\pi}{10/3} = \dfrac{6\pi}{10} \approx 1.885 \text{ rad/m} \)
Step 3: Calculate peak magnetic field using \( E_0 = c B_0 \).
\( B_0 = \dfrac{E_0}{c} = \dfrac{0.05}{3 \times 10^8} \approx 1.67 \times 10^{-10} \text{ T} \)
Step 4: Verify \( c = \omega / k \).
\( \dfrac{\omega}{k} = \dfrac{5.655 \times 10^8}{1.885} \approx 3 \times 10^8 \text{ m/s} \) ✓
Answer
(a) \( \omega \approx 5.655 \times 10^8 \) rad/s (b) \( k \approx 1.885 \) rad/m (c) \( B_0 \approx 1.67 \times 10^{-10} \) T
CBSE Exam Tips 2025-26
- Memorise the speed of light: Always use \( c = 3 \times 10^8 \) m/s in calculations. We recommend writing this value at the top of your answer sheet during the exam.
- Convert units before substituting: Frequencies are often given in MHz or GHz. Convert to Hz first. For example, 100 MHz = \( 10^8 \) Hz.
- Know the EM spectrum order: CBSE 2025-26 papers frequently ask for the position of radio waves in the electromagnetic spectrum. Radio waves have the lowest frequency and longest wavelength.
- State the formula clearly: In board exams, always write the formula \( c = f\lambda \) before substituting values. This earns step marks even if the final answer has an arithmetic error.
- Learn Planck’s constant: \( h = 6.626 \times 10^{-34} \) J·s is given in the CBSE data sheet, but knowing it saves time. Use it for photon energy questions.
- Practice dimensional analysis: Verify your answers by checking units. Frequency (Hz = s⁻¹) times wavelength (m) must give m/s.
Common Mistakes to Avoid
- Forgetting to convert frequency units: A very common error is substituting MHz directly into the formula without converting to Hz. Always multiply MHz by \( 10^6 \) and GHz by \( 10^9 \) before using \( c = f\lambda \).
- Confusing wavelength and frequency: Radio waves have high wavelength and low frequency compared to other EM waves. Students often reverse this, especially when comparing the EM spectrum.
- Using the wrong value of Planck’s constant: Some students use \( h = 6.6 \times 10^{-34} \) J/Hz instead of J·s. The unit is joule-seconds, not joules per hertz (though they are equivalent, writing the unit correctly matters in board exams).
- Assuming radio waves travel slower in air: For exam purposes, radio waves travel at \( c = 3 \times 10^8 \) m/s in both vacuum and air. The difference is negligible and NCERT treats them as equal.
- Mixing up \(\omega\) and \(f\): Angular frequency \( \omega \) is in rad/s, while frequency \( f \) is in Hz. The relation \( \omega = 2\pi f \) is essential. Never substitute \( f \) where \( \omega \) is required in a wave equation.
JEE/NEET Application of Radio Waves Formula
In our experience, JEE aspirants and NEET students encounter radio waves and the electromagnetic spectrum in several predictable question patterns. Mastering the Radio Waves Formula gives you a clear advantage in these sections.
Pattern 1: Electromagnetic Spectrum Classification
JEE Main frequently asks students to arrange EM waves in order of frequency or wavelength. Radio waves occupy the range from 3 Hz to 300 GHz. Knowing that \( \lambda = c/f \) allows you to calculate and compare wavelengths instantly. In JEE 2023 and 2024, at least one question per paper involved identifying which EM wave corresponds to a given frequency range.
Pattern 2: Energy and Momentum of Radio Photons
NEET often tests the energy of photons across the EM spectrum. Since radio waves have the lowest frequency, they also carry the least energy per photon. Use \( E = hf \) to calculate and compare. A common NEET question asks: “Which EM wave has the lowest photon energy?” — the answer is always radio waves. Numerically, for a 1 MHz radio wave, \( E = 6.626 \times 10^{-28} \) J, which is far smaller than visible light photons (\( \sim 10^{-19} \) J).
Pattern 3: Wave Parameters in Oscillating Fields
JEE Advanced problems on electromagnetic waves require calculating \( \omega \), \( k \), \( E_0 \), and \( B_0 \) together. The key relationships are \( \omega = 2\pi f \), \( k = 2\pi/\lambda \), \( c = \omega/k \), and \( E_0 = cB_0 \). Our experts suggest practising problems where two of these are given and the rest must be derived — this is the standard JEE Advanced format. The intensity formula \( I = \frac{1}{2}c\varepsilon_0 E_0^2 \) also appears in JEE Advanced and links electromagnetic wave theory with energy concepts.
For NEET, focus on conceptual understanding: radio waves are used in radar (detection), AM/FM broadcasting, and satellite communication. These applications are directly mentioned in NCERT Class 12 and appear as one-mark conceptual questions in NEET.
FAQs on Radio Waves Formula
Explore More Physics Formulas
Understanding the Radio Waves Formula becomes even more powerful when studied alongside related topics. We recommend exploring the Induced Voltage Formula, which explains how changing electromagnetic fields produce voltage — a concept directly linked to how radio antennas work. You should also study the Superposition Formula to understand how radio waves combine and interfere. For a broader understanding of wave mechanics, the Critical Velocity Formula provides useful context on wave propagation in different media.
For a complete collection of all Physics formulas covered in NCERT Class 12, visit our Physics Formulas hub page. You can also refer to the official NCERT website at ncert.nic.in for the original Class 12 Physics textbook content on electromagnetic waves.