de Broglie Wavelength: Formula and How to Calculate It
Chemistry · Structure Of Atom · NEET
Every moving object has a wavelength. de Broglie said λ = h/mv = h/p, where h is Planck's constant (6.626 × 10⁻³⁴ J·s), m is mass, and v is velocity. So a fast, light particle (like an electron) has a longer, measurable wavelength; a heavy object has a tiny one. Memory hook: "Little mass, long wave" — small m on the bottom makes λ big.
λ = h/mv: mass sits in the denominator, so a light electron gets a long, detectable wave while a heavy cricket ball gets a wavelength too small to measure.
Your doubts, answered
What exactly is the de Broglie wavelength formula?
The formula is λ = h/mv, which is the same as λ = h/p because momentum p = mv. Here λ is the wavelength (in metres), h = 6.626 × 10⁻³⁴ J·s is Planck's constant, m is mass in kg, and v is velocity in m/s. This one equation connects a particle (mass, velocity) to a wave (wavelength). NEET loves it because it links matter and waves in a single line.
Is de Broglie wavelength h/mv or h/p? They look different.
They are the SAME. Momentum p is defined as p = mv, so h/p and h/mv are identical. Use h/mv when the question gives you mass and velocity. Use h/p when the question directly gives momentum. Do not get confused — just remember p = mv and both forms match.
What units do I put in so the answer comes out in metres?
Always use SI units: mass in kilograms (kg), velocity in metres per second (m/s), and h in J·s. Then λ comes out in metres (m). If mass is given in grams, divide by 1000 to get kg first. A common NEET mistake is leaving mass in grams — that makes the answer 1000 times wrong.
How do I find the wavelength when the question gives kinetic energy instead of velocity?
Use λ = h/√(2mKE). This comes from KE = ½mv², so p = mv = √(2m·KE). Just plug mass (kg) and kinetic energy (in joules) into λ = h/√(2m·KE). This form saves time when velocity is not given directly.
What if an electron is accelerated through a voltage V — what wavelength does it get?
Its kinetic energy becomes KE = eV (charge × voltage). So λ = h/√(2meV). For an electron a handy shortcut is λ = 12.27/√V ångström (where V is in volts). This is common in NEET/JEE for electron microscope questions.
Why is the de Broglie wavelength so tiny for a cricket ball but big for an electron?
Because mass is on the bottom of λ = h/mv. A cricket ball has a huge mass, so λ becomes unimaginably small (around 10⁻³⁴ m) and cannot be measured. An electron has a very tiny mass, so its λ is large enough to be detected (electron diffraction proves this). Remember: small mass → long wave.
Does a stationary (not moving) object have a de Broglie wavelength?
No. If v = 0, then mv = 0, and λ = h/0 becomes infinite — meaning the concept has no physical wave. de Broglie wavelength only applies to a particle in motion. This is a favourite NTA trap: 'A particle at rest has a de Broglie wavelength' is FALSE.
⚠️ The NEET trap ✗ de Broglie wavelength is λ = h·mv (mass and velocity multiplied on top), or you can leave mass in grams. ✓ λ = h/mv = h/p — mass and velocity are in the DENOMINATOR, and mass must be in kg (SI units). Larger mass or velocity gives a SMALLER wavelength. 🧠 h is on top, mv is on the bottom. 'Heavy and fast → short wave.' If your answer grows when mass grows, you flipped the formula.
Real NEET questions
NEET 2017 / 2018
Which one is the wrong statement?
A · de Broglie's wavelength is given by λ = h/mv, where m = mass of the particle and v = group velocity of the particle
B · The uncertainty principle is ΔE · Δt ≥ h/4π
C · Half-filled and fully filled orbitals have greater stability due to greater exchange energy, greater symmetry and more balanced arrangement
D · The energy of the 2s orbital is less than the energy of the 2p orbital in case of hydrogen-like atoms ✓
Solution: The de Broglie relation λ = h/mv is correct, so option A is a TRUE statement (not the answer). The wrong statement is D: in single-electron (hydrogen-like) species, orbital energy depends only on n, so 2s and 2p are degenerate (equal energy). This PYQ confirms the de Broglie formula λ = h/mv is examined directly by NTA.
NEET 2019 (Odisha)
In a hydrogen atom, the de Broglie wavelength of an electron in the second Bohr orbit is: [Given Bohr radius a₀ = 52.9 pm]
A · 211.6 pm
B · 211.6π pm ✓
C · 52.9π pm
D · 105.8 pm
Solution: Bohr's condition gives nλ = 2πrₙ. The radius of the nth orbit is rₙ = a₀·n²/Z = 52.9 × (2)²/1 = 211.6 pm. So λ = 2πrₙ/n = (2π × 211.6)/2 = 211.6π pm. Answer B. This links the de Broglie wavelength directly to the Bohr orbit — the next concept in this chain.
Solved Structure Of Atom NEET PYQs
Try the real previous-year questions from this chapter — each with the answer and a full solution.
What is the de Broglie wavelength formula for an electron?
For any particle, including an electron, λ = h/mv = h/p. Put the electron mass (9.11 × 10⁻³¹ kg), its velocity in m/s, and h = 6.626 × 10⁻³⁴ J·s to get λ in metres.
What is the SI unit of de Broglie wavelength?
The metre (m). In atomic problems the answer is often written in nanometres (nm), picometres (pm), or ångström (Å): 1 Å = 10⁻¹⁰ m.
Can we calculate de Broglie wavelength from kinetic energy?
Yes. Since p = √(2m·KE), the wavelength is λ = h/√(2m·KE). Use mass in kg and kinetic energy in joules.
Why can't we see the de Broglie wavelength of everyday objects?
Their mass is very large, so λ = h/mv becomes about 10⁻³⁴ m — far too small to measure. Only very light particles like electrons have a detectable wavelength.
Is the de Broglie wavelength directly or inversely proportional to velocity?
Inversely. Because v is in the denominator (λ = h/mv), higher velocity gives a shorter wavelength, and lower velocity gives a longer wavelength.
