Amino acids can be classified as $\alpha$-, $\beta$-, $\gamma$-, $\delta$- and so on depending upon the relative position of the amino group with respect to the carboxyl group. Which type of amino acids forms polypeptide chain in proteins?

Amino acids can be classified as α-, β-, γ-, δ- and so on depending upon the relative position of the amino gr

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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.

\lambda = \frac{h}{p} = \frac{h}{mv}

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:

\lambda = \frac{6.626 \times 10^{-34}}{0.16 \times 30} \approx 1.38 \times 10^{-34} \ \text{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 $9.1 \times 10^{-31}$ kg) accelerated through 100 volts. Its speed is about $5.9 \times 10^6$ m/s. Its de Broglie wavelength:

\lambda = \frac{6.626 \times 10^{-34}}{9.1 \times 10^{-31} \times 5.9 \times 10^6} \approx 1.23 \times 10^{-10} \ \text{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.

Note

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 $\lambda = \frac{h}{p}$ .

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: $p = \frac{h}{\lambda}$ (from $E = pc$ for light, combined with $c = f\lambda$ )

So for light:

\lambda = \frac{h}{p}

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 $p$ :

\boxed{\lambda = \frac{h}{p}}

Where:

$\lambda$ = de Broglie wavelength
$h$ = Planck's constant ( $6.626 \times 10^{-34} \text{ J·s}$ )
$p$ = 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):

\lambda = \frac{h}{mv}

For a relativistic particle (like an electron at high speed):

p = \gamma mv \quad \text{where} \quad \gamma = \frac{1}{\sqrt{1 - v^2/c^2}}

\lambda = \frac{h}{\gamma mv}

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: $f = \frac{E}{h}$
The wavelength comes from momentum: $\lambda = \frac{h}{p}$

These two relations are linked by the phase velocity of the wave:

v_{\text{phase}} = f\lambda = \frac{E}{h} \cdot \frac{h}{p} = \frac{E}{p}

For a free particle with kinetic energy $E = \frac{p^2}{2m}$ :

v_{\text{phase}} = \frac{p}{2m} = \frac{v}{2}

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: …

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Only $\alpha$ -amino acids — those with the amino group on the carbon adjacent to the carboxyl group — form the polypeptide chains found in proteins, because their structure allows peptide bonds to link without steric hindrance.

The key to this question lies in understanding what a polypeptide chain actually is, and why nature chose a specific geometry for it.

Proteins are built from amino acids linked by peptide bonds. A peptide bond forms between the carboxyl group ( $-COOH$ ) of one amino acid and the amino group ( $-NH_2$ ) of another. But here’s the catch: for this bond to form efficiently and for the resulting chain to fold into functional shapes, the amino and carboxyl groups must be attached to the same carbon atom — the $\alpha$ -carbon.

Why? Because in an $\alpha$ -amino acid, the amino group and the carboxyl group are separated by exactly one carbon. This places them close enough to react easily, and the resulting backbone has a regular, repeating pattern: $—NH—CHR—CO—$ . This pattern is what allows the chain to adopt stable secondary structures like $\alpha$ -helices and $\beta$ -sheets.

If you used a $\beta$ -amino acid (amino group on the second carbon from the carboxyl), the backbone would have an extra carbon inserted, making the chain longer, more flexible, and unable to pack into the compact, stable folds required for protein function. $\gamma$ -, $\delta$ -, and higher amino acids would make the backbone even more irregular.

So the answer is straightforward: only $\alpha$ -amino acids are used in natural protein synthesis.

Watch out

A common mistake is to think that any amino acid with both groups can form a peptide bond. While technically possible in a lab, only $\alpha$ -amino acids produce the specific backbone geometry that living systems use for functional proteins.

Tip

The $\alpha$ -carbon is the central carbon in an amino acid. It’s the one that carries the amino group, the carboxyl group, a hydrogen atom, and a variable side chain (R group). If the amino group is on any other carbon, it’s not an $\alpha$ -amino acid.

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The Core Concept

Amino acids in proteins are linked by peptide bonds to form polypeptide chains. The key point is that only $\alpha$ -amino acids (where the amino group is attached to the carbon adjacent to the carboxyl group) are used in natural protein synthesis.

Common Mistakes & How to Avoid Them

✗ Mistake 1: Thinking any amino acid can form a polypeptide chain

Why it happens: Students memorise that "amino acids form proteins" but forget the structural restriction.

How to avoid:

Remember: only $\alpha$ -amino acids have the amino group on the $\alpha$ -carbon (the carbon next to the –COOH group).
In $\beta$ -, $\gamma$ -, or $\delta$ -amino acids, the amino group is further away — this prevents the correct alignment needed for peptide bond formation during ribosomal translation.

Key exam point: Polypeptide chains in proteins are built exclusively from $\alpha$ -amino acids.

✗ Mistake 2: Confusing "polypeptide chain" with "peptide bond formation in general"

Why it happens: Some non- $\alpha$ amino acids can form amide bonds in lab conditions, so students assume they can form polypeptides in nature.

How to avoid:

Distinguish between chemical peptide bond formation (possible with any amino acid in a test tube) and biological polypeptide synthesis (ribosome-based, requires $\alpha$ -amino acids).
In exams, "polypeptide chain in proteins" always refers to natural, biological chains.

✗ Mistake 3: Forgetting the $\alpha$ -carbon definition

Why it happens: Students confuse the numbering of carbon atoms in amino acids.

How to avoid:

Draw a generic amino acid: $\text{H}_2\text{N} - \text{CHR} - \text{COOH}$ The carbon bearing both the –NH $_2$ and –COOH groups is the $\alpha$ -carbon.
If the amino group is on the next carbon (C2 from –COOH), it's $\beta$ ; on C3, it's $\gamma$ , and so on.

✗ Mistake 4: Assuming $\beta$ - or $\gamma$ -amino acids are irrelevant

Why it happens: Some textbooks mention them only briefly, so students ignore them.

How to avoid: …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

KCET 2026Set D31 markMCQ
$\lambda = \dfrac{h}{\sqrt{m\nu}}$
›Reveal solutionSolution

The de Broglie relation connects a moving particle's wavelength to its momentum via Planck's constant.

Step 1 — de Broglie's hypothesis

Louis de Broglie proposed that every moving particle of mass $m$ and velocity $\nu$ has an associated "matter wave" whose wavelength $\lambda$ is inversely proportional to its momentum $p = m\nu$ .

Step 2 — The equation

This gives $\lambda = \dfrac{h}{p} = \dfrac{h}{m\nu}$ , where $h$ is Planck's constant.

Step 3 — Eliminating the other options …
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KCET 2023Set A-31 markMCQ
$Q. When light propagates through a given homogeneous medium, the velocities of (A) primary wavefronts are lesser than those of secondary wavelets. (B) primary wavefronts are greater than or equal to those of secondary wavelets. (C) primary wavefront and wavelets are equal. (D) primary wavefront are larger than those of secondary wavelets.$
›Reveal solutionSolution
Huygens' principle: in a homogeneous medium the secondary wavelets travel at the same speed as the primary wavefront, because the medium offers the same speed everywhere.

1. The concept — Huygens' principle.

Huygens' geometrical construction says:

Every point on a primary wavefront behaves as a fresh source of disturbance, emitting secondary wavelets.

These wavelets travel in the forward direction with the speed of the wave in that medium.

After a time $t$ , the new position of the wavefront is the forward envelope (common tangent) of all the secondary wavelets.

2. Why the two speeds must be identical here.

The new wavefront is constructed as the envelope of wavelets of radius $vt$ . If the wavelets moved at some speed different from $v$ , the envelope they build would advance at that different speed — the wavefront's speed simply is the wavelets' speed. In a homogeneous medium the speed $v = c/n$ is the same at every point and in every direction, so:
$v_{\text{wavelet}} = v_{\text{wavefront}} = \frac{c}{n}$
3. Why the other options fail. …
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KCET 2022Set B-31 markMCQ
$Q. “Heat cannot be itself flow from a body at lower temperature to a body at higher temperature”. This statement corresponds to (A) Conservation of mass (B) First law of thermodynamics (C) Second law of Thermodynamics (D) Conservation of momentum$
›Reveal solutionSolution

The quoted sentence is verbatim the Clausius statement of the Second Law of Thermodynamics.

Step 1 — Recognise the statement.

The sentence "heat cannot of itself flow from a body at lower temperature to a body at higher temperature" is the classic Clausius statement of the Second Law of Thermodynamics. The two crucial words are "of itself" — meaning spontaneously, with no external agency.

Step 2 — Why the First Law cannot be the answer.

The First Law is just energy conservation:
$\Delta Q = \Delta U + \Delta W$
It is completely indifferent to direction. If $100\ \mathrm{J}$ flowed spontaneously from a cold body to a hot one, energy would still be perfectly conserved — the First Law would be entirely satisfied. Yet such a process is never observed. So the First Law is not sufficient to explain the one-way nature of heat flow; something more is needed, and that something is the Second Law.

Step 3 — What the Second Law adds: direction.

The Second Law supplies the arrow of time for thermal processes, quantified by entropy: for any spontaneous process in an isolated system,
$\Delta S_{\text{total}} \geq 0$
If heat $Q$ passed spontaneously from a cold reservoir at $T_C$ to a hot one at $T_H$ (with $T_H > T_C$ ):

$\Delta S_{\text{total}} = \underbrace{-\frac{Q}{T_C}}_{\text{cold body loses}} + \underbrace{\frac{Q}{T_H}}_{\text{hot body gains}} = Q\left(\frac{1}{T_H} - \frac{1}{T_C}\right) < 0$ …
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KCET 2022Set B-31 markMCQ
$2.2 \times 10^{-11}m$
›Reveal solutionSolution

Use $p = h/\lambda$ — the de Broglie / photon-momentum relation — and divide.

1. Why $p = h/\lambda$

A photon has zero rest mass, so its momentum cannot be found from $p = mv$ . From special relativity, a massless particle satisfies $E = pc$ ; and from Planck's hypothesis its energy is $E = h\nu$ . Combining, and using $c = \nu\lambda$ :
$p = \frac{E}{c} = \frac{h\nu}{c} = \frac{h\nu}{\nu\lambda}$ $\boxed{\,p = \frac{h}{\lambda}\,}$
(This is also exactly de Broglie's relation, read backwards — it is the same equation that assigns a wavelength to matter.)

Notice that $c$ has cancelled: we do not need the speed of light, which is why the question only supplies $h$ and $\lambda$ .

2. Substitute
$h = 6.6\times10^{-34}\ \text{J s}, \qquad \lambda = 2.2\times10^{-11}\ \text{m}$ $p = \frac{6.6\times10^{-34}}{2.2\times10^{-11}}$
3. Arithmetic

Mantissas: $\dfrac{6.6}{2.2} = 3$

Exponents: $10^{-34-(-11)} = 10^{-23}$ …
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KCET 2021Set B-21 markMCQ
$10^\circ$
›Reveal solutionSolution
For a pendulum to execute simple harmonic motion, the restoring torque must be directly proportional to the angular displacement. This requires both a small bob (so it behaves as a point mass) and a small angular amplitude (so $\sin\theta \approx \theta$ ). The correct option is (A).

The key idea is that simple harmonic motion (SHM) arises only when the restoring force (or torque) is proportional to the displacement from equilibrium. For a pendulum, the restoring torque is $mgL\sin\theta$ , where $\theta$ is the angular displacement. This is proportional to $\sin\theta$ , not $\theta$ itself. The approximation $\sin\theta \approx \theta$ (in radians) holds only for small angles — typically less than about $10^\circ$ . That’s why condition (II) is necessary.

But why does the bob size matter? The formula for the period $T = 2\pi\sqrt{L/g}$ assumes the pendulum is a simple pendulum: a point mass suspended by a massless, inextensible string. If the bob is large, its size is not negligible compared to the length $L$ , and the pendulum behaves as a physical pendulum — the centre of mass shifts, and the moment of inertia changes. The motion is still oscillatory, but the simple harmonic approximation fails unless the bob is small enough to treat as a point mass. So condition (I) is also necessary.

Let’s go through the reasoning step by step.

The condition for SHM

For any oscillating system, SHM requires a restoring force (or torque) that is directly proportional to the displacement, and opposite in direction. For a pendulum, the restoring torque about the pivot is $\tau = -mgL\sin\theta$ . For small $\theta$ , $\sin\theta \approx \theta$ (in radians), so $\tau \approx -mgL\theta$ . This is of the form $\tau = -k\theta$ , which gives SHM. The approximation is accurate to within about 1% for $\theta < 10^\circ$ (about 0.175 rad). So condition (II) is essential.

Why the bob size matters …
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KCET 2020Set A-11 markMCQ
$e^-$
›Reveal solutionSolution
In the photoelectric effect, the number of photoelectrons ejected per second is directly proportional to the intensity of incident light (for a fixed frequency above threshold), while the kinetic energy of each electron depends only on frequency and work function — not on intensity.

The photoelectric effect is a beautiful demonstration of the particle nature of light. When light of sufficient frequency (above the threshold frequency) strikes a metal surface, it ejects electrons. The key insight is that light behaves as a stream of photons, each carrying energy $h\nu$ . One photon interacts with one electron, transferring its entire energy. If that energy exceeds the work function $\phi$ (the minimum energy needed to free the electron), the electron is ejected with kinetic energy $K_{\text{max}} = h\nu - \phi$ .

This one-photon–one-electron rule is the foundation. It tells us that the number of ejected electrons depends on how many photons strike the surface per second — that is, the intensity. The energy of each ejected electron depends on the photon energy $h\nu$ , not on how many photons arrive.

Let’s examine each option carefully.

Option (A): Number of $e^-$ ejected increases with the increase in the intensity of incident light.

Intensity is energy per unit area per second. For monochromatic light, intensity $I = n h\nu$ , where $n$ is the number of photons per second per unit area. If you increase $I$ while keeping frequency $\nu$ fixed, $n$ increases. More photons mean more electrons ejected (provided $\nu$ is above threshold). This is correct.

Option (B): Energy of $e^-$ ejected increases with the increase in the intensity of incident light.

The maximum kinetic energy of an ejected electron is $K_{\text{max}} = h\nu - \phi$ . Intensity does not appear here. Changing intensity changes the number of photons, not the energy per photon. So the energy of each electron remains the same. This is false.

Option (C): Number of $e^-$ ejected increases with the increase in the frequency of incident light. …
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Q.Amino acids can be classified as $\alpha$ -, $\beta$ -, $\gamma$ -, $\delta$ - and so on depending upon the relative position of the amino group with respect to the carboxyl group. Which type of amino acids forms polypeptide chain in proteins?

Concept understanding — De Broglie Wavelength

De Broglie Wavelength: When Particles Start Acting Like Waves

The Core Idea

Why You Don't See It in Daily Life

What This Means Physically

Why this formula?

De Broglie Wavelength: Why the Formula Holds

The Core Insight: Nature's Symmetry

Step 1: Start with Light (What We Already Knew)

Step 2: De Broglie's Bold Hypothesis

Step 3: Why Momentum and Not Velocity?

Step 4: The Deeper Reasoning — Wave-Particle Duality

Step 5: Experimental Confirmation (Why We Believe It)

Concept: Amino Acid Classification by Position of Amino Group

Method: Structural Identification & Biological Context

Steps:

Answer

Why only $\alpha$ -amino acids? …

The Core Concept

Common Mistakes & How to Avoid Them

✗ Mistake 1: Thinking any amino acid can form a polypeptide chain

✗ Mistake 2: Confusing "polypeptide chain" with "peptide bond formation in general"

✗ Mistake 3: Forgetting the $\alpha$ -carbon definition

✗ Mistake 4: Assuming $\beta$ - or $\gamma$ -amino acids are irrelevant

Step 1 — de Broglie's hypothesis

Step 2 — The equation

Step 3 — Eliminating the other options …

1. Why $p = h/\lambda$

2. Substitute

3. Arithmetic

Q.Amino acids can be classified as α\alphaα-, β\betaβ-, γ\gammaγ-, δ\deltaδ- and so on depending upon the relative position of the amino group with respect to the carboxyl group. Which type of amino acids forms polypeptide chain in proteins?

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Q.Amino acids can be classified as $\alpha$ -, $\beta$ -, $\gamma$ -, $\delta$ - and so on depending upon the relative position of the amino group with respect to the carboxyl group. Which type of amino acids forms polypeptide chain in proteins?

Unlock everything free for 14 days