Q.Some enzymes are named after the reaction, where they are used. What name is given to the class of enzymes which catalyse the oxidation of one substrate with simultaneous reduction of another substrate?
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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: …
The key idea here is the classification of enzymes based on the type of reaction they catalyse. The question describes a reaction where one substrate is oxidised (loses electrons) while another is simultaneously reduced (gains electrons). This is a coupled oxidation-reduction reaction, which is the defining feature of a specific enzyme class. …
The class of enzymes that catalyse the oxidation of one substrate while simultaneously reducing another is called oxidoreductases. This is the key concept: they transfer electrons (or hydrogen atoms) between two molecules, coupling oxidation and reduction in a single reaction.
Enzymes are biological catalysts, and their naming often reflects the reaction they speed up. When you see a reaction where one molecule loses electrons (gets oxidised) and another gains those electrons (gets reduced), you are looking at a redox reaction. The enzyme that makes this happen must handle both halves of the electron transfer.
The logic is straightforward: the name of the enzyme class comes directly from the type of reaction it catalyses. Since the reaction involves both oxidation and reduction, the name combines these two ideas.
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Identify the reaction type. The question describes a reaction where one substrate is oxidised and another is simultaneously reduced. This is a redox reaction (reduction-oxidation). In such a reaction, electrons (or hydrogen atoms) are transferred from the donor (which gets oxidised) to the acceptor (which gets reduced).
-
Match the reaction to the enzyme class. The International Union of Biochemistry and Molecular Biology (IUBMB) classifies enzymes into six major classes based on the reaction they catalyse. The class that handles all redox reactions is Class 1: Oxidoreductases. …
Concept: Enzyme Classification by Reaction Type
Enzymes are classified into six major classes by the International Union of Biochemistry and Molecular Biology (IUBMB), based on the type of reaction they catalyse.
The class you are asking about is:
Oxidoreductases
Method: IUBMB Enzyme Classification System
Step 1 — Identify the reaction type
- The reaction involves oxidation of one substrate (loss of electrons or hydrogen) and reduction of another substrate (gain of electrons or hydrogen).
- This is a redox (reduction-oxidation) reaction.
Step 2 — Match to the enzyme class
- Enzymes that catalyse redox reactions are grouped under Class 1: Oxidoreductases.
- They often have names ending in -dehydrogenase, -oxidase, -reductase, or -oxygenase, depending on the specific electron acceptor.
Step 3 — Confirm the definition …
Here is the breakdown of the common mistakes students make on this specific concept, along with how to avoid them.
The Core Concept: Enzyme Classification & Oxidoreductases
The question asks for the class of enzymes that catalyze oxidation-reduction (redox) reactions where one substrate is oxidized (loses electrons/H) and another is reduced (gains electrons/H).
The correct answer is Oxidoreductases.
Common Mistake #1: Confusing the Class with a Sub-class
- The Mistake: Students often answer with a specific type of oxidoreductase (like Dehydrogenase, Oxidase, or Reductase) instead of the broad Class name.
- Why it happens: The question mentions "oxidation" and "reduction," so students jump to the most familiar term (e.g., "dehydrogenase" for removing hydrogen).
- How to Avoid It:
- Remember the Hierarchy: The IUBMB (International Union of Biochemistry and Molecular Biology) classifies enzymes into 6 main classes. The class is the broadest category.
- Memorize the 6 Classes:
- Oxidoreductases (Redox reactions)
- Transferases (Transfer functional groups)
- Hydrolases (Hydrolysis)
- Lyases (Addition/removal of groups without hydrolysis)
- Isomerases (Isomerization)
- Ligases (Joining two molecules using ATP)
- Key Distinction: The question asks for the class. "Dehydrogenase" is a sub-class within Oxidoreductases. Always read if the question asks for "class," "sub-class," or "type."
Common Mistake #2: Confusing Oxidoreductases with Transferases
- The Mistake: Students answer Transferases because they see the phrase "transfer of electrons" or "transfer of hydrogen" and think of it as a group transfer.
- How to Avoid It:
- Understand the Definition: Transferases transfer functional groups (e.g., methyl, amino, phosphate groups) from one molecule to another. They do not involve a change in oxidation state.
- The Redox Rule: If the reaction involves a change in the oxidation number of atoms (gain/loss of electrons or hydrogen atoms), it is always an Oxidoreductase, not a Transferase.
- Example: The transfer of a phosphate group from ATP to glucose (catalyzed by Hexokinase) is a Transferase. The removal of hydrogen from lactate to form pyruvate (catalyzed by Lactate Dehydrogenase) is an Oxidoreductase.
Common Mistake #3: Forgetting the "Simultaneous" Nature
- The Mistake: Students answer Oxidase or Oxygenase, thinking only of oxidation.
- Why it happens: The question explicitly says "oxidation of one substrate with simultaneous reduction of another substrate." Students focus only on the "oxidation" part.
- How to Avoid It:
- The "Coupling" Rule: In biology, oxidation and reduction are always coupled. You cannot have one without the other. …
- 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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