Q.Find the ratio of the de Broglie wavelengths associated with an alpha particle and a proton, if both
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🔒 Start your 14-day free trial to unlock the full solution →Concept understanding — De Broglie Wavelength
De Broglie Wavelength: When Particles Start Acting Like Waves
Imagine you're holding a cricket ball. You know exactly where it is, and if you throw it, you can predict its path. That's a particle — localised, definite, following Newton's laws. Now think of light. You can't "hold" a beam of light; it spreads out, bends around corners, creates interference patterns. That's a wave — spread out, not localised.
For centuries, physics kept these two worlds separate. Particles were particles. Waves were waves. Never the twain shall meet.
Then came a young French physicist, Louis de Broglie, in 1924. He asked a question that seemed almost absurd: If light — which we thought was a wave — can behave like a particle (the photoelectric effect), then why can't a particle — say, an electron — behave like a wave?
That question turned physics upside down.
The Core Idea
De Broglie proposed that every moving particle has a wave associated with it. The wavelength of that wave depends on the particle's momentum. The faster or heavier the particle, the shorter the wavelength.
λ=ph=mvh
Where:
- λ = de Broglie wavelength (in metres)
- h = Planck's constant (6.626×10−34 J⋅s)
- p = momentum of the particle (mv for non-relativistic speeds)
This is not a mathematical trick. It's a physical reality. An electron moving through a crystal actually behaves like a wave of this wavelength — it can diffract, interfere, and form patterns just like light does.
Why You Don't See It in Daily Life
Here's the crucial point: the de Broglie wavelength is incredibly tiny for everyday objects.
Take a cricket ball of mass 0.16 kg moving at 30 m/s. Its de Broglie wavelength is:
λ=0.16×306.626×10−34≈1.38×10−34 m
That's about a hundred trillion trillion times smaller than the nucleus of an atom. No experiment can detect such a wave — it's effectively zero for all practical purposes.
Now take an electron (mass 9.1×10−31 kg) accelerated through 100 volts. Its speed is about 5.9×106 m/s. Its de Broglie wavelength:
λ=9.1×10−31×5.9×1066.626×10−34≈1.23×10−10 m
That's about 0.12 nanometres — comparable to the spacing between atoms in a crystal. This is measurable. And indeed, in 1927, Davisson and Germer fired electrons at a nickel crystal and observed diffraction — the unmistakable signature of a wave.
The de Broglie wavelength is only observable when it is comparable to the size of objects the particle interacts with. For macroscopic objects, it's far too small to matter. For subatomic particles, it's the key to understanding their behaviour.
What This Means Physically
The wave is not a physical wave in space like a water wave. It's a probability wave — its amplitude at any point tells you the probability of finding the particle there. Where the wave amplitude is large, you're likely to find the particle; where it's zero, you won't.
This wave-particle duality is not a compromise. It's the actual nature of reality. An electron is neither a pure particle nor a pure wave — it's something that shows particle-like behaviour in some experiments (like hitting a screen at a point) and wave-like behaviour in others (like passing through two slits and interfering with itself). …
Why this formula?
De Broglie Wavelength: Why the Formula Holds
Let's build this from the ground up — understanding why matter has a wavelength, not just memorizing λ=ph.
The Core Insight: Nature's Symmetry
Before de Broglie, physics had two separate worlds:
- Light — showed wave behaviour (diffraction, interference) but also particle behaviour (photoelectric effect)
- Matter — showed particle behaviour (momentum, collisions) but no wave behaviour yet
De Broglie asked a daring question in his 1924 PhD thesis:
If light (a wave) can behave like a particle, why can't a particle (like an electron) behave like a wave?
Nature should be symmetric — what applies to one should apply to the other.
Step 1: Start with Light (What We Already Knew)
For a photon, Einstein had given us two key relations:
- Energy: E=hf (Planck's relation)
- Momentum: p=λh (from E=pc for light, combined with c=fλ)
So for light:
λ=ph
This was experimentally verified for photons.
Step 2: De Broglie's Bold Hypothesis
De Broglie said: This relation is not special to light. It is universal.
For any particle with momentum p:
λ=ph
Where:
- λ = de Broglie wavelength
- h = Planck's constant (6.626×10−34 J⋅s)
- p = momentum of the particle
Step 3: Why Momentum and Not Velocity?
This is crucial. The formula uses momentum (p=mv), not just velocity.
For a non-relativistic particle (slow compared to light):
λ=mvh
For a relativistic particle (like an electron at high speed):
p=γmvwhereγ=1−v2/c21
λ=γmvh
Why momentum? Because momentum is the more fundamental quantity — it's conserved, it's frame-independent in a deeper sense, and it connects directly to the wave's phase.
Step 4: The Deeper Reasoning — Wave-Particle Duality
De Broglie didn't just guess. He reasoned:
- Every moving particle has an associated wave — called the "matter wave" or "pilot wave"
- The frequency of this wave comes from energy: f=hE
- The wavelength comes from momentum: λ=ph
These two relations are linked by the phase velocity of the wave:
vphase=fλ=hE⋅ph=pE
For a free particle with kinetic energy E=2mp2:
vphase=2mp=2v
This is half the particle's speed — a strange but mathematically consistent result.
Step 5: Experimental Confirmation (Why We Believe It)
De Broglie's idea was confirmed when electrons showed wave behaviour: …
Concept: de Broglie wavelength λ=ph=mvh, where momentum p=mv for non-relativistic particles.
Key data: Mass of alpha particle mα=4mp (since it contains 2 protons + 2 neutrons), charge qα=2e; proton has mass mp, charge e.
(a) Same speeds v:
λpλα=mαvmpv=4mpmp=41
(b) Same kinetic energy K:
Since K=21mv2, we have v=m2K, so λ=2mKh.
λpλα=mαmp=4mpmp=21
(c) Same potential difference V: …
The de Broglie wavelength λ=ph depends on momentum. For particles with the same speed, wavelength ratios equal inverse mass ratios; for the same kinetic energy or potential, they depend on m1. The ratios are (a) 1:4, (b) 1:2, (c) 1:22.
The de Broglie wavelength connects the wave and particle nature of matter through the relation λ=ph, where h is Planck's constant and p is momentum. Since momentum depends on both mass and velocity (or equivalently on mass and kinetic energy), the wavelength ratio between two particles will change depending on what physical quantity they share in common.
An alpha particle consists of 2 protons and 2 neutrons, so its mass is approximately mα=4mp and its charge is qα=2e. The proton has mass mp and charge qp=e.
(a) Same speeds
When both particles move at the same speed v, their momenta differ only by their masses.
- Write the de Broglie wavelengths:
λα=mαvh,λp=mpvh
- Form the ratio: λpλα=mαvmpv=mαmp=4mpmp=41 …
- GUJCET 2026Set x1 markMCQQ.The relation between the wavelength of electromagnetic radiation (λ) and de Broglie wavelength of its quantum (photon) (λ′) is ______. (A) λ′>λ (B) λ′=λ (C) λ′<λ (D) λ′=2λ
›Reveal solutionSolution
For a photon the de Broglie wavelength equals the EM wavelength: λ′=λ.
A photon of electromagnetic radiation of wavelength λ carries momentum
p=λh
Its de Broglie wavelength is
λ′=ph=h/λh=λ …
- GSEB Higher Secondary Certificate (HSC) Examination 2026Set ANNUAL1 markMCQQ.The de Broglie wavelength of proton and alpha-particle is same. The ratio of their velocities is ___.(a) 4 : 1(b) 1 : 2(c) 2 : 1(d) 1 : 4
›Reveal solutionSolution
Equal de Broglie wavelengths mean equal momenta (lambda = h/p), so the lighter particle must move faster in inverse proportion to the mass ratio.
lambda = h/(m v), so equal lambda means m_p v_p = m_alpha v_alpha.
…
- GSEB Higher Secondary Certificate (HSC) Examination 2026Set ANNUAL1 markMCQQ.A proton, a neutron, an electron and an alpha-particle have same energy. Then their de-Broglie wavelengths compare as(a) lambda_e = lambda_p = lambda_n = lambda_a(b) lambda_e < lambda_p = lambda_n > lambda_a(c) lambda_a < lambda_p = lambda_n < lambda_e(d) lambda_p = lambda_n > lambda_e > lambda_a
›Reveal solutionSolution
At the same kinetic energy E, de Broglie wavelength lambda = h/sqrt(2 m E) is inversely proportional to sqrt(mass), so the heaviest particle has the smallest wavelength.
lambda = h / sqrt(2 m E). For fixed E, lambda ~ 1/sqrt(m).
Masses: m_e (electron) is by far the smallest; m_p (proton) approx equals m_n (neutron); m_alpha (alpha particle) approx 4 m_p, the largest.
So, ordering by increasing mass: electron < proton = neutron < alpha. …
- GSEB Higher Secondary Certificate (HSC) Examination 2024Set ANNUAL1 markMCQQ.If de-Broglie wavelength of a dust particle of mass 1.0 x 10^-9 kg is 3 x 10^-25 m then the speed of the particle is ___. (h = 6.625 x 10^-34 Js)(a) 1.1 ms^-1(b) 1.0 kms^-1(c) 1.2 kms^-1(d) 2.2 ms^-1
›Reveal solutionSolution
de Broglie wavelength: λ = h/(mv), so v = h/(mλ).
m = 1.0 × 10⁻⁹ kg, λ = 3 × 10⁻²⁵ m, h = 6.625 × 10⁻³⁴ Js.
…
- GSEB Higher Secondary Certificate (HSC) Examination 2023Set ANNUAL1 markMCQQ.The de Broglie wavelength (lambda) associated with an electron accelerated through a potential difference of 121 V is ___. [m_e = 9.1 x 10^-31 kg, h = 6.63 x 10^-34 Js](a) 12.0 A(b) 2.1 A(c) 1.12 A(d) 0.12 A
›Reveal solutionSolution
For an electron accelerated through V volts, lambda = 12.27/sqrt(V) angstrom; with V = 121 V this is 1.12 A.
The de Broglie wavelength of an electron accelerated through potential difference V is: …
- GUJCET 2019Set 131 markMCQQ.To increase de Broglie wavelength of an electron from 0.5×10−10 m to 10−10 m, its energy should be............. (A) Decreased to fourth part (B) Doubled (C) Halved (D) Increased to 4 times
›Reveal solutionSolution
λ∝E−1/2, so doubling λ requires E reduced to one-fourth.
Concept: de Broglie wavelength λ=2mEh∝E1, therefore E∝λ21.
Steps:
- λ increases by factor 0.5×10−1010−10=2. …
- GSEB Higher Secondary Certificate (HSC) Examination 2018Set ANNUAL1 markMCQQ.The uncertainty in position of a particle is same as it's de Broglie wavelength, uncertainty in its momentum is ___.(a) h/lambda(b) 2h/3lambda(c) lambda/h(d) 3lambda/2h
›Reveal solutionSolution
Using Heisenberg's relation delta_x . delta_p approximately h with delta_x = lambda gives delta_p = h/lambda.
Heisenberg's uncertainty principle (in the simple form): delta_x . delta_p approximately h.
Given the uncertainty in position equals the de Broglie wavelength, delta_x = lambda.
Then delta_p approximately h / delta_x = h / lambda.
…
- GUJCET 2015Set C1 markMCQQ.If alpha particle and deutron move with velocity v and 2v respectively, the ratio of their de-Broglie wave length will be _____. (A) 2:1 (B) 1:2 (C) 1:1 (D) 2:1
›Reveal solutionSolution
[!TLDR]
λα=h/(4uv) and λd=h/(4uv) are equal, so λα:λd=1:1. Answer: (C).
Concept
The de Broglie wavelength of a particle is λ=mvh (NCERT/CBSE dual nature of matter). An alpha particle has mass ≈4u; a deuteron has mass ≈2u.
Solution
For the alpha particle (mass 4u, speed v):
λα=(4u)(v)h=4uvh …
- GUJCET 2015Set C1 markMCQQ.de-Broglie wave length of atom at TK absolute temperature will be (A) 3mKTh (B) mKTh (C) h2mKT (D) 2mKT
›Reveal solutionSolution
[!TLDR] λ=h/p with p=3mKT gives λ=3mKTh.
Concept
A particle in thermal equilibrium at temperature T has average translational kinetic energy KE=23kT (k = Boltzmann constant). Its momentum is p=2mKE, and the de-Broglie wavelength is λ=h/p.
Solution
KE=23kT …
- GUJCET 2014Set A1 markMCQQ.If the kinetic energy of free electron is made double, the new de Broglie wave length will be __________ times that of initial wave length. (A) 2 (B) 21 (C) 2 (D) 21
›Reveal solutionSolution
[!TLDR]
The new wavelength is 21 times the original.
Concept
For a particle of mass m and kinetic energy E, momentum p=2mE, so the de Broglie wavelength is
λ=ph=2mEh∝E1.
Solution
If E→2E (mass unchanged): …
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