The De Broglie Wavelength Formula, expressed as \( \lambda = \frac{h}{mv} \), describes the wave nature of matter and is a cornerstone of modern physics studied in NCERT Class 12 Physics, Chapter 11 (Dual Nature of Radiation and Matter). This formula is essential for CBSE board exams and appears frequently in JEE Main, JEE Advanced, and NEET. In this article, we cover the complete derivation, a comprehensive formula sheet, three progressive solved examples, CBSE exam tips, common mistakes, and JEE/NEET applications.
Key De Broglie Wavelength Formulas at a Glance
Quick reference for the most important De Broglie wavelength formulas.
- Basic De Broglie relation: \( \lambda = \frac{h}{mv} \)
- In terms of momentum: \( \lambda = \frac{h}{p} \)
- For accelerated particle: \( \lambda = \frac{h}{\sqrt{2mK}} \)
- For electron accelerated through potential V: \( \lambda = \frac{h}{\sqrt{2meV}} \)
- Thermal De Broglie wavelength: \( \lambda = \frac{h}{\sqrt{3mk_BT}} \)
- Relation to kinetic energy: \( \lambda = \frac{h}{\sqrt{2mE_k}} \)
- Photon wavelength (Einstein): \( E = hf = \frac{hc}{\lambda} \)
What is De Broglie Wavelength Formula?
The De Broglie Wavelength Formula was proposed by French physicist Louis de Broglie in 1924. He suggested that every moving particle with mass and velocity possesses wave-like properties. This revolutionary idea extended the concept of wave-particle duality from photons to all matter. The formula states that the wavelength associated with a moving particle is inversely proportional to its momentum.
This concept is covered in NCERT Class 12 Physics, Chapter 11 (Dual Nature of Radiation and Matter). It is also relevant to Class 11 students exploring modern physics basics. The De Broglie Wavelength Formula bridges classical mechanics and quantum mechanics. It explains phenomena such as electron diffraction and is the foundation of quantum wave mechanics developed by Schrödinger. Understanding this formula is critical for any student aiming to score well in CBSE board exams or crack JEE/NEET.
The formula is mathematically stated as:
\[ \lambda = \frac{h}{p} = \frac{h}{mv} \]
Here, \( \lambda \) is the de Broglie wavelength, \( h \) is Planck's constant, \( m \) is the mass of the particle, and \( v \) is its velocity.
De Broglie Wavelength Formula — Expression and Variables
The standard expression for the De Broglie Wavelength Formula is:
\[ \lambda = \frac{h}{mv} \]
| Symbol | Quantity | SI Unit |
|---|---|---|
| \( \lambda \) | De Broglie wavelength | Metre (m) |
| \( h \) | Planck's constant | Joule-second (J·s), value = 6.626 × 10⁻³⁴ J·s |
| \( m \) | Mass of the particle | Kilogram (kg) |
| \( v \) | Velocity of the particle | Metre per second (m/s) |
| \( p \) | Momentum of the particle (p = mv) | kg·m/s |
| \( K \) or \( E_k \) | Kinetic energy of the particle | Joule (J) |
| \( V \) | Accelerating potential (for charged particles) | Volt (V) |
| \( e \) | Charge of electron | Coulomb (C), value = 1.6 × 10⁻¹⁹ C |
| \( k_B \) | Boltzmann constant | J/K, value = 1.38 × 10⁻²³ J/K |
| \( T \) | Absolute temperature | Kelvin (K) |
Derivation of De Broglie Wavelength Formula
De Broglie combined Einstein's mass-energy relation and Planck's quantum theory to derive this formula. The derivation proceeds as follows:
Step 1: From Einstein's energy-mass relation, the energy of a photon is \( E = mc^2 \).
Step 2: From Planck's quantum theory, the energy of a photon is \( E = hf = \frac{hc}{\lambda} \).
Step 3: Equating both expressions: \( mc^2 = \frac{hc}{\lambda} \), which gives \( \lambda = \frac{h}{mc} \).
Step 4: De Broglie extended this to any particle moving with velocity \( v \) by replacing \( c \) with \( v \) and \( mc \) with momentum \( p = mv \):
\[ \lambda = \frac{h}{mv} = \frac{h}{p} \]
Step 5: For a particle with kinetic energy \( K = \frac{1}{2}mv^2 \), we get \( p = \sqrt{2mK} \), so:
\[ \lambda = \frac{h}{\sqrt{2mK}} \]
This derivation elegantly shows that every particle in motion has an associated wavelength. Heavier or faster particles have shorter wavelengths. This is why macroscopic objects show no observable wave nature, while electrons exhibit clear diffraction patterns.
Complete Physics Formula Sheet — Dual Nature and Quantum Physics
| Formula Name | Expression | Variables | SI Units | NCERT Chapter |
|---|---|---|---|---|
| De Broglie Wavelength (basic) | \( \lambda = h/mv \) | h = Planck's constant, m = mass, v = velocity | m | Class 12, Ch 11 |
| De Broglie (momentum form) | \( \lambda = h/p \) | p = momentum | m | Class 12, Ch 11 |
| De Broglie (kinetic energy) | \( \lambda = h/\sqrt{2mK} \) | K = kinetic energy | m | Class 12, Ch 11 |
| De Broglie (accelerated electron) | \( \lambda = h/\sqrt{2meV} \) | e = electron charge, V = potential | m | Class 12, Ch 11 |
| Thermal De Broglie wavelength | \( \lambda = h/\sqrt{3mk_BT} \) | k_B = Boltzmann constant, T = temperature | m | Class 12, Ch 11 |
| Planck's Energy Relation | \( E = hf \) | f = frequency | J | Class 12, Ch 11 |
| Photoelectric Effect (Einstein) | \( KE_{max} = hf – \phi \) | φ = work function | J | Class 12, Ch 11 |
| Work Function | \( \phi = hf_0 \) | f₀ = threshold frequency | J or eV | Class 12, Ch 11 |
| Stopping Potential | \( eV_0 = KE_{max} \) | V₀ = stopping potential | V | Class 12, Ch 11 |
| Bohr's Quantisation Condition | \( mvr = n\hbar \) | r = orbit radius, n = quantum number | kg·m²/s | Class 12, Ch 12 |
| Heisenberg Uncertainty Principle | \( \Delta x \cdot \Delta p \geq h/4\pi \) | Δx = position uncertainty, Δp = momentum uncertainty | J·s | Class 12, Ch 11 |
De Broglie Wavelength Formula — Solved Examples
Example 1 (Class 10-11 Level)
Problem: Calculate the de Broglie wavelength of a ball of mass 0.1 kg moving with a velocity of 10 m/s. (Planck's constant h = 6.626 × 10⁻³⁴ J·s)
Given: m = 0.1 kg, v = 10 m/s, h = 6.626 × 10⁻³⁴ J·s
Step 1: Write the De Broglie Wavelength Formula: \( \lambda = \frac{h}{mv} \)
Step 2: Calculate momentum: \( p = mv = 0.1 \times 10 = 1 \) kg·m/s
Step 3: Substitute values: \( \lambda = \frac{6.626 \times 10^{-34}}{1} = 6.626 \times 10^{-34} \) m
Answer
The de Broglie wavelength of the ball is \( 6.626 \times 10^{-34} \) m. This extremely small value explains why macroscopic objects do not exhibit observable wave behaviour.
Example 2 (Class 12 Level)
Problem: An electron (mass = 9.11 × 10⁻³¹ kg) is accelerated through a potential difference of 100 V. Calculate its de Broglie wavelength. (h = 6.626 × 10⁻³⁴ J·s, e = 1.6 × 10⁻¹⁹ C)
Given: m = 9.11 × 10⁻³¹ kg, V = 100 V, h = 6.626 × 10⁻³⁴ J·s, e = 1.6 × 10⁻¹⁹ C
Step 1: Use the formula for an accelerated particle: \( \lambda = \frac{h}{\sqrt{2meV}} \)
Step 2: Calculate \( 2meV \):
\( 2meV = 2 \times 9.11 \times 10^{-31} \times 1.6 \times 10^{-19} \times 100 \)
\( = 2 \times 9.11 \times 1.6 \times 10^{-48} = 29.152 \times 10^{-48} \) kg²·m²/s²
Step 3: Calculate \( \sqrt{2meV} \):
\( \sqrt{29.152 \times 10^{-48}} = 5.399 \times 10^{-24} \) kg·m/s
Step 4: Calculate \( \lambda \):
\( \lambda = \frac{6.626 \times 10^{-34}}{5.399 \times 10^{-24}} = 1.227 \times 10^{-10} \) m = 1.227 Å
Answer
The de Broglie wavelength of the electron is approximately \( 1.227 \times 10^{-10} \) m (1.227 Å). This is comparable to X-ray wavelengths, which is why electrons can exhibit diffraction.
Example 3 (JEE/NEET Level)
Problem: A proton and an alpha particle are accelerated through the same potential difference V. Find the ratio of their de Broglie wavelengths \( \lambda_p / \lambda_{\alpha} \). (Mass of proton = m, charge = e; mass of alpha = 4m, charge = 2e)
Given: Both particles accelerated through same potential V. For proton: mass = m, charge = e. For alpha particle: mass = 4m, charge = 2e.
Step 1: Write the formula for each particle:
\( \lambda = \frac{h}{\sqrt{2 \cdot M \cdot q \cdot V}} \) where M = mass and q = charge of the particle.
Step 2: For the proton:
\( \lambda_p = \frac{h}{\sqrt{2meV}} \)
Step 3: For the alpha particle:
\( \lambda_{\alpha} = \frac{h}{\sqrt{2 \cdot 4m \cdot 2e \cdot V}} = \frac{h}{\sqrt{16meV}} \)
Step 4: Find the ratio:
\( \frac{\lambda_p}{\lambda_{\alpha}} = \frac{h/\sqrt{2meV}}{h/\sqrt{16meV}} = \frac{\sqrt{16meV}}{\sqrt{2meV}} = \sqrt{\frac{16}{2}} = \sqrt{8} = 2\sqrt{2} \)
Answer
The ratio of de Broglie wavelengths \( \lambda_p / \lambda_{\alpha} = 2\sqrt{2} \approx 2.83 \). The proton has a longer de Broglie wavelength than the alpha particle when accelerated through the same potential. This type of ratio problem is a standard JEE question pattern.
CBSE Exam Tips 2025-26
- Memorise all formula variants: The basic form \( \lambda = h/mv \), the kinetic energy form \( \lambda = h/\sqrt{2mK} \), and the accelerated-particle form \( \lambda = h/\sqrt{2meV} \) are all asked in CBSE 2025-26 board exams. We recommend practising all three in one sitting.
- Value of Planck's constant: Always use h = 6.626 × 10⁻³⁴ J·s (or 6.63 × 10⁻³⁴ J·s). Write the value clearly in your answer sheet. Examiners deduct marks for missing constants.
- Unit conversion: Convert the final wavelength to Ångströms (1 Å = 10⁻¹⁰ m) or nanometres (1 nm = 10⁻⁹ m) when the question asks. CBSE 2025-26 marking schemes often expect the unit in Å for atomic-scale problems.
- Derivation questions: The derivation of the De Broglie Wavelength Formula is a 3-mark question in CBSE. Write all steps clearly, starting from Einstein's energy relation and Planck's quantum hypothesis.
- Conceptual questions: Be prepared to explain why macroscopic objects do not show wave nature. The answer lies in the extremely small value of \( \lambda \) for large masses. This is a common 2-mark question in CBSE 2025-26.
- Link to Bohr's model: Our experts suggest connecting the De Broglie Wavelength Formula to Bohr's quantisation condition \( mvr = n\hbar \). This shows that standing electron waves explain stable orbits — a frequent 5-mark essay question.
Common Mistakes to Avoid
- Using wrong mass units: Many students use mass in grams instead of kilograms. Always convert mass to kg before substituting in \( \lambda = h/mv \). For example, an electron mass is 9.11 × 10⁻³¹ kg, not 9.11 × 10⁻²⁸ g.
- Confusing kinetic energy with potential energy: When an electron is accelerated through potential V, its kinetic energy is K = eV (not just V). Always write \( K = qV \) first, then substitute into \( \lambda = h/\sqrt{2mK} \).
- Forgetting the square root in the energy formula: A very common error is writing \( \lambda = h/(2mK) \) instead of \( \lambda = h/\sqrt{2mK} \). The momentum is \( p = \sqrt{2mK} \), not \( 2mK \).
- Applying the formula to photons incorrectly: The De Broglie Wavelength Formula uses rest mass. Photons have zero rest mass, so the formula \( \lambda = h/mv \) does not directly apply. For photons, use \( \lambda = hc/E \) instead.
- Ignoring significant figures: CBSE and JEE both require answers to 2-3 significant figures. Do not write an answer like 1.22699 × 10⁻¹⁰ m — round it to 1.23 × 10⁻¹⁰ m.
JEE/NEET Application of De Broglie Wavelength Formula
In our experience, JEE aspirants encounter the De Broglie Wavelength Formula in 1-2 questions per paper, typically in the Modern Physics section. NEET also tests this formula in the Dual Nature of Matter chapter. Here are the three most common application patterns:
Pattern 1: Ratio of Wavelengths for Different Particles
JEE frequently asks for the ratio of de Broglie wavelengths of two particles (proton vs. electron, proton vs. alpha, etc.) accelerated through the same or different potentials. The key relation is:
\[ \frac{\lambda_1}{\lambda_2} = \sqrt{\frac{m_2 q_2 V_2}{m_1 q_1 V_1}} \]
Practise this pattern extensively. It appears in JEE Main almost every year.
Pattern 2: De Broglie Wavelength and Bohr's Orbit
JEE Advanced often combines the De Broglie Wavelength Formula with Bohr's atomic model. The condition for a standing wave in a Bohr orbit is:
\[ 2\pi r_n = n\lambda \]
This means the circumference of the nth orbit equals n times the de Broglie wavelength. Substituting \( \lambda = h/mv \) gives Bohr's quantisation condition \( mvr = nh/2\pi \). This elegant connection is a high-value topic in JEE Advanced.
Pattern 3: Thermal De Broglie Wavelength
For NEET and JEE Advanced, the thermal de Broglie wavelength of a gas molecule at temperature T is tested. The formula is:
\[ \lambda_{thermal} = \frac{h}{\sqrt{3mk_BT}} \]
This arises because the average kinetic energy of a gas molecule is \( \frac{3}{2}k_BT \). Questions ask how \( \lambda \) changes with temperature — as T increases, \( \lambda \) decreases. Our experts note that this pattern tests conceptual understanding rather than pure calculation, making it a favourite in NEET MCQs.
We also recommend revising the Electric Field Formula and the Uniform Circular Motion Formula alongside the De Broglie Wavelength Formula, as JEE questions often combine these topics in multi-concept problems.
FAQs on De Broglie Wavelength Formula
Explore More Physics Formulas
Now that you have mastered the De Broglie Wavelength Formula, we recommend exploring related topics to build a complete understanding of modern physics and electromagnetism. Visit our Physics Formulas hub for a complete list of Class 11 and Class 12 formulas. You may also find the following articles helpful:
- Electric Field Formula — essential for understanding force on charged particles, directly related to electron acceleration problems.
- Uniform Circular Motion Formula — used alongside Bohr's model to derive orbital velocities and connect with de Broglie wavelengths.
- Capacitance Formula — important for understanding potential difference in accelerating electrons.
- Refractive Index Formula — connects wave optics with the dual nature of light, complementing your study of wave-particle duality.
For official NCERT resources and syllabus details, visit the NCERT official website.