Describe the term D- and L-configuration used for amino acids with examples.

NCERT Exemplar · Q46

Q. Describe the term D- and L-configuration used for amino acids with examples.

CBSEShort· 3mImportance★★★★★

72% · 79/110 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

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.

\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: …

Start your 14-day free trial to unlock the full solution →

D- and L-configuration in amino acids refers to the absolute spatial arrangement around the chiral $\alpha$ -carbon, determined by the Fischer projection relative to glyceraldehyde. L-amino acids have the amino group on the left in the standard Fischer projection; D-amino acids have it on the right. Almost all natural amino acids are L.

The Core Idea: Chirality and the Glyceraldehyde Reference

Amino acids (except glycine) have a chiral $\alpha$ -carbon — four different groups attached: an amino group ( $-\text{NH}_2$ ), a carboxyl group ( $-\text{COOH}$ ), a hydrogen atom ( $-\text{H}$ ), and a variable side chain ( $-\text{R}$ ). This carbon can exist in two mirror-image forms (enantiomers). The D/L system is an older but still widely used naming convention that links the configuration of any chiral molecule back to the reference standard: glyceraldehyde.

\text{L-glyceraldehyde: } \text{OH on left, CH}_2\text{OH at top, CHO at bottom}

The trick is to draw the amino acid in a Fischer projection with the carboxyl group ( $-\text{COOH}$ ) at the top and the side chain ( $-\text{R}$ ) at the bottom. Then, if the amino group ( $-\text{NH}_2$ ) is on the left, it is L; if on the right, it is D.

Step-by-Step: How to Assign D or L

Draw the Fischer projection. Place the most oxidised carbon (the carboxyl carbon, $-\text{COOH}$ ) at the top. Place the side chain ( $-\text{R}$ ) at the bottom. The chiral $\alpha$ -carbon is at the intersection of the vertical and horizontal lines. The hydrogen ( $-\text{H}$ ) and amino group ( $-\text{NH}_2$ ) occupy the two horizontal positions.
Check the position of the $-\text{NH}_2$ group. This is the only step that matters for the D/L label.
- If $-\text{NH}_2$ is on the left horizontal arm → L-configuration.
- If $-\text{NH}_2$ is on the right horizontal arm → D-configuration.
Important: This is NOT the same as optical rotation. The D/L label is about absolute configuration (spatial arrangement), not the direction in which the molecule rotates plane-polarised light (which is denoted by $+$ or $-$ ). An L-amino acid can be dextrorotatory ( $+$ ) or levorotatory ( $-$ ) depending on its side chain.

Watch out

A common mistake is to confuse D/L with the R/S system (Cahn-Ingold-Prelog). They are not equivalent. For example, L-alanine is (S)-alanine, but L-cysteine is (R)-cysteine because the sulfur atom in the side chain changes the priority order. Always use the Fischer projection method for D/L.

Examples

L-Alanine (the natural form)

Structure: $\text{CH}_3-\text{CH}(\text{NH}_2)-\text{COOH}$
Fischer projection: COOH at top, $\text{CH}_3$ at bottom. $\text{NH}_2$ on the left, H on the right.
Result: $\text{NH}_2$ is left → L-Alanine.
Note: L-Alanine is also (S)-alanine.

D-Alanine (found in bacterial cell walls)

Structure: Same molecular formula, mirror image.
Fischer projection: COOH at top, $\text{CH}_3$ at bottom. $\text{NH}_2$ on the right, H on the left.
Result: $\text{NH}_2$ is right → D-Alanine.
Note: D-Alanine is (R)-alanine.

L-Cysteine (a special case)

Structure: $\text{HS}-\text{CH}_2-\text{CH}(\text{NH}_2)-\text{COOH}$ …

Start your 14-day free trial to unlock the full solution →

Concept: Stereochemistry of Amino Acids – D and L Configuration

The D/L system is a relative configuration system used to describe the spatial arrangement of atoms around the chiral carbon (the $\alpha$ -carbon) in amino acids. It is based on the reference molecule glyceraldehyde.

Method: The Fischer Projection Rule (Glyceraldehyde Reference)

Name of method: Fischer Projection Comparison Method

Steps:

Draw the amino acid in a Fischer projection
- Place the $-\text{COOH}$ group at the top.
- Place the $-\text{R}$ group (side chain) at the bottom.
- The $-\text{NH}_2$ group and $-\text{H}$ are on the left and right sides.
Identify the position of the $-\text{NH}_2$ group
- If $-\text{NH}_2$ is on the left → L-configuration (like L-glyceraldehyde).
- If $-\text{NH}_2$ is on the right → D-configuration (like D-glyceraldehyde).
Compare with the reference
- In L-glyceraldehyde, the $-\text{OH}$ is on the left.
- In L-amino acids, the $-\text{NH}_2$ is on the left (analogous position).

Examples

Amino Acid	Fischer Projection (simplified)	Configuration
Alanine	$\text{COOH}$ at top, $\text{CH}_3$ at bottom, $\text{NH}_2$ on left	L-Alanine
Alanine	$\text{COOH}$ at top, $\text{CH}_3$ at bottom, $\text{NH}_2$ on right	D-Alanine

Start your 14-day free trial to unlock the full solution →

Common Mistakes: D- and L-Configuration in Amino Acids

Students often struggle with this concept because it involves spatial orientation and historical naming conventions that differ from modern R/S notation. Here are the most frequent errors and how to avoid them.

Mistake 1: Confusing D/L with R/S

The error: Students think D = R and L = S, or vice versa.

Why it's wrong: D/L is based on the Fischer projection of glyceraldehyde (a reference standard), while R/S is based on the Cahn-Ingold-Prelog priority rules. They are not equivalent — for example, L-cysteine has the R configuration at the $\alpha$ -carbon, not S.

How to avoid:

Remember: D/L is absolute configuration relative to glyceraldehyde, not a direct priority-based assignment.
For most amino acids (except cysteine), L = S and D = R, but always check the specific molecule.
Use the mnemonic: "L = Left" — in the Fischer projection of an L-amino acid, the amino group ( $-NH_2$ ) is on the left.

Mistake 2: Forgetting the Reference Standard

The error: Students define D/L without mentioning glyceraldehyde as the reference.

Why it's wrong: The D/L system was historically defined using (+)-glyceraldehyde and (−)-glyceraldehyde. An amino acid is D if its $\alpha$ -carbon configuration matches that of D-glyceraldehyde.

How to avoid:

Always state: "The D/L configuration is assigned by comparing the configuration of the $\alpha$ -carbon to that of D- or L-glyceraldehyde."
Draw the Fischer projection of glyceraldehyde side-by-side with the amino acid.

Mistake 3: Misidentifying the Chiral Carbon

The error: Students apply D/L to the wrong carbon (e.g., a side-chain chiral center).

Why it's wrong: D/L refers only to the $\alpha$ -carbon (the carbon bearing both the amino and carboxyl groups). Side-chain chiral centers (e.g., in threonine or isoleucine) are described separately.

How to avoid:

Circle the $\alpha$ -carbon in every amino acid structure before assigning D/L.
For amino acids with multiple chiral centers (like threonine), use the three-letter prefix (e.g., L-threonine refers only to the $\alpha$ -carbon; the side-chain is described as threo or erythro).

Mistake 4: Mixing Up Fischer Projection Rules

The error: Students incorrectly draw the Fischer projection — placing the amino group on the wrong side.

Why it's wrong: In a standard Fischer projection of an L-amino acid:

The carboxyl group ( $-COOH$ ) is at the top.
The side chain (R group) is at the bottom.
The amino group ( $-NH_2$ ) is on the left.
The hydrogen ( $-H$ ) is on the right.

How to avoid:

Memorize the "COHN" mnemonic: Carboxyl at top, O (oxygen) not needed, H on right, N on left for L.
Practice drawing both D and L forms of alanine (simplest chiral amino acid) until it's automatic.

Mistake 5: Forgetting That Glycine Has No D/L

The error: Students assign D or L to glycine.

Why it's wrong: Glycine ( $H_2N-CH_2-COOH$ ) has no chiral carbon — its $\alpha$ -carbon has two hydrogen atoms. Therefore, it is achiral and has no D/L configuration.

How to avoid:

Before assigning D/L, check: "Does this amino acid have four different groups attached to the $\alpha$ -carbon?"
If the answer is no (glycine), write "not applicable" or "achiral."

Mistake 6: Confusing D/L with Optical Rotation (+/−)

The error: Students think D = dextrorotatory (clockwise) and L = levorotatory (anticlockwise). …

Start your 14-day free trial to unlock the full solution →

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

Showing the 12 most recent of 70 on this concept.

CBSE 2026Set 55/1/11 markMCQ
$Q.For questions 13 to 16, two statements are given – one labelled Assertion (A) and the other labelled Reason (R). Select the correct answer from the codes (A), (B), (C) and (D) below: (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A). (B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A). (C) Assertion (A) is true, but Reason (R) is false. (D) Both Assertion (A) and Reason (R) are false. Assertion (A) : If accelerated electrons are passed through a narrow slit, a diffraction pattern is observed. Reason (R) : Electrons behave as both particles and waves.$
›Reveal solutionSolution

Electrons exhibit wave-particle duality; their wave nature (de Broglie wavelength) causes diffraction when passing through narrow slits, just like light waves. Both statements are true, and the Reason correctly explains the Assertion.

Understanding Wave-Particle Duality

The heart of this question lies in one of quantum mechanics' most profound insights: matter at the atomic scale doesn't fit neatly into our classical categories of "particle" or "wave." Louis de Broglie proposed in 1924 that every moving particle has an associated wavelength, given by
$\lambda = \frac{h}{p} = \frac{h}{mv}$
where $h$ is Planck's constant, $p$ is momentum, $m$ is mass, and $v$ is velocity.

When we accelerate electrons, we increase their momentum. Yet even at high speeds, electrons retain a measurable de Broglie wavelength—typically on the order of angstroms for electrons accelerated through a few hundred volts. This wavelength is comparable to the spacing between atoms in crystals or the width of carefully engineered slits.

Examining the Assertion

Assertion (A): If accelerated electrons are passed through a narrow slit, a diffraction pattern is observed.

This is experimentally verified and true. The classic demonstration is the Davisson-Germer experiment (1927), which showed electron diffraction from crystal lattices. More dramatically, modern versions of the double-slit experiment with electrons—sending them one at a time—build up an interference pattern on a detector screen over time.

Diffraction occurs when waves encounter obstacles or apertures comparable to their wavelength. The electron beam, despite being composed of particles with mass and charge, produces the characteristic bright and dark fringes we associate with wave phenomena. The central maximum, secondary maxima, and minima all appear exactly as wave theory predicts.
$\theta$
Examining the Reason

Reason (R): Electrons behave as both particles and waves.

This is the principle of wave-particle duality, a cornerstone of quantum mechanics. It is unequivocally true.

Electrons exhibit particle properties: they have definite mass ( $9.11 \times 10^{-31}$ kg), charge ( $-1.6 \times 10^{-19}$ C), and produce localized impacts on detectors (you can count individual electron arrivals). Simultaneously, they exhibit wave properties: they diffract, interfere, and possess a wavelength and frequency.

Neither description alone is complete. The electron is a quantum object, and which aspect we observe depends on the experimental setup. When we look for particle behavior (measuring position or momentum), we find particles. When we create conditions for wave behavior (slits, crystals), we observe diffraction and interference. …
Start your 14-day free trial to see the full solution →
CBSE 2026Set 55/3/11 markMCQ
$\left(\dfrac{E_p}{E_\alpha}\right)$
›Reveal solutionSolution
For equal momentum, kinetic energy is inversely proportional to mass, and de Broglie wavelength is directly proportional to mass. Since the alpha particle has 4 times the mass of a proton, the ratio of kinetic energies is $4:1$ and the ratio of wavelengths is $1:1$ . The correct option is (C).

The key to this problem lies in two fundamental relationships: the de Broglie wavelength and the connection between kinetic energy and momentum. When two particles have the same momentum, their de Broglie wavelengths become equal — that part is immediate. The kinetic energy, however, depends on mass because $E_k = p^2/2m$ , so the lighter particle has more kinetic energy.

Let’s work through it systematically.

Recall the de Broglie wavelength formula.

Every moving particle has a wavelength associated with it, given by $\lambda = \frac{h}{p}$ , where $h$ is Planck’s constant and $p$ is the linear momentum. This is a direct consequence of wave-particle duality — the more momentum a particle has, the shorter its wavelength.

Since the problem states that the proton and alpha particle have equal momentum, we can write:

$p_p = p_\alpha$
Therefore:
$\lambda_p = \frac{h}{p_p} \quad \text{and} \quad \lambda_\alpha = \frac{h}{p_\alpha}$
Because $p_p = p_\alpha$ , the two wavelengths are identical:
$\frac{\lambda_p}{\lambda_\alpha} = 1$

Now find the kinetic energy ratio. Kinetic energy is related to momentum by:

$E_k = \frac{p^2}{2m}$
This comes from combining $E_k = \frac12 mv^2$ with $p = mv$ . For equal momentum, the kinetic energy is inversely proportional to mass — a heavier particle moving with the same momentum must be slower, so it has less kinetic energy.

For the proton (mass $m_p$ ) and alpha particle (mass $m_\alpha$ ):
$\frac{E_p}{E_\alpha} = \frac{p^2 / 2m_p}{p^2 / 2m_\alpha} = \frac{m_\alpha}{m_p}$

Know the masses involved. …
Start your 14-day free trial to see the full solution →
CBSE 2026Set V11 markMCQ
$\alpha$
(a) proton
(b) electron
(c) neutron
(d) $\alpha$ -particle
›Reveal solutionSolution

(b) electron …

Start your 14-day free trial to see the full solution →
CBSE 2026Set A1 markMCQ
$Q. The ratio of de Broglie wavelength associated with two electrons accelerated through 49 V and 64 V is (A) 49/64 (B) 64/49 (C) 7/8 (D) 8/7$
›Reveal solutionSolution

de Broglie wavelength of an accelerated electron λ ∝ 1/√V; ratio = √(64/49) = 8/7.

For an electron accelerated through potential difference V, its kinetic energy is $eV = \frac{p^2}{2m}$ , so $p = \sqrt{2meV}$ and the de Broglie wavelength is

$\lambda = \frac{h}{\sqrt{2meV}} \propto \frac{1}{\sqrt{V}}.$

Hence …

Start your 14-day free trial to see the full solution →
CBSE 2026Set ANNUAL1 markMCQ
$Q. If alpha particle, proton and electron move with the same momentum, then their respective de Broglie wavelengths λα, λp, λe are related as$
(a) λα > λp > λe
(b) λα < λp < λe
(c) λα = λp = λe
(d) None of the above
›Reveal solutionSolution

Equal momentum means equal de Broglie wavelength, regardless of the particles' different masses.

The de Broglie wavelength of a particle is

$\lambda = \dfrac{h}{p}$

where $h$ is Planck's constant and $p$ is the particle's momentum. This formula depends only on momentum $p$ , not on the particle's mass, charge, or identity. Since the alpha particle, proton and electron are all stated to have the same momentum, they must all have the …

Start your 14-day free trial to see the full solution →
CBSE 2026Set ANNUAL1 mark
$Q. A body of mass 0.10 kg is moving with a speed of 10 m/s. The de-Broglie wavelength of the wave associated with it will be ______ m.$
›Reveal solutionSolution

The de Broglie wavelength of a moving body is lambda = h/(mv); plugging in the mass and speed gives the value directly.

De Broglie's relation: lambda = h / (m v), where h = 6.63 x 10^-34 J.s (Planck's constant), m = 0.10 kg, v = 10 m/s. …

Start your 14-day free trial to see the full solution →
CBSE 2025Set 55/4/11 markMCQ
$Q.Choose the correct statement: (A) Photons of light show diffraction whereas electrons do not show diffraction. (B) Electrons have momentum whereas photons do not have momentum. (C) Photons of light and electrons both exhibit dual nature. (D) All electromagnetic radiations do not have photons.$
›Reveal solutionSolution
Both photons and electrons exhibit wave-particle duality — the key idea is that both show diffraction (wave behaviour) and carry momentum (particle behaviour). The correct statement is (C).

Concept First: De Broglie Wavelength and Dual Nature

The entire foundation of modern quantum mechanics rests on wave-particle duality — the idea that every moving particle has a wavelength associated with it. Louis de Broglie proposed this in 1924, and it’s summarised by:
$\lambda = \frac{h}{p}$
where $\lambda$ is the de Broglie wavelength, $h$ is Planck’s constant, and $p$ is the momentum.

This means:

Photons (light quanta) have momentum $p = \frac{h}{\lambda}$ and show wave phenomena like diffraction and interference.

Electrons (particles with mass) also have a de Broglie wavelength $\lambda = \frac{h}{mv}$ and therefore do show diffraction — this was famously confirmed by Davisson and Germer in 1927.

So both photons and electrons possess both wave-like and particle-like properties. That is the dual nature.

Step-by-Step Analysis

1. Statement (A): “Photons of light show diffraction whereas electrons do not show diffraction.”

This is false. Electrons do show diffraction — the Davisson–Germer experiment proved it. In fact, electron diffraction is routinely used in techniques like transmission electron microscopy (TEM). The wavelength of an electron can be tuned by changing its accelerating voltage, making it a practical tool.

Watch out

A common mistake is to think diffraction is only for light. In reality, any particle with a de Broglie wavelength comparable to the slit spacing will diffract — electrons, neutrons, even large molecules like buckyballs have been shown to diffract.

2. Statement (B): “Electrons have momentum whereas photons do not have momentum.” …
Start your 14-day free trial to see the full solution →
CBSE 2025Set 55/5/11 markMCQ
$\dfrac{\lambda_{\alpha}}{\lambda_{p}}$
›Reveal solutionSolution
The de Broglie wavelength depends on momentum, not directly on kinetic energy. Using $K = \frac{p^2}{2m}$ and the given $K_\alpha = 4K_p$ , along with $m_\alpha = 4m_p$ , we find $\frac{\lambda_\alpha}{\lambda_p} = \frac{1}{4}$ , which corresponds to option (C).

The de Broglie wavelength is the bridge between particle and wave behaviour: $\lambda = \frac{h}{p}$ , where $h$ is Planck’s constant and $p$ is the momentum. The problem gives you kinetic energy, not momentum directly — so the first step is always to connect $K$ and $p$ .

For any non-relativistic particle, kinetic energy is $K = \frac{p^2}{2m}$ . Rearranging, $p = \sqrt{2mK}$ . This is the key relation that lets you translate the given energy ratio into a wavelength ratio.

Write the wavelength for each particle.

For the alpha particle: $\lambda_\alpha = \frac{h}{p_\alpha} = \frac{h}{\sqrt{2 m_\alpha K_\alpha}}$ .

For the proton: $\lambda_p = \frac{h}{p_p} = \frac{h}{\sqrt{2 m_p K_p}}$ .

Take the ratio.

$\frac{\lambda_\alpha}{\lambda_p} = \frac{h / \sqrt{2 m_\alpha K_\alpha}}{h / \sqrt{2 m_p K_p}} = \sqrt{\frac{m_p K_p}{m_\alpha K_\alpha}}.$
Notice that $h$ and the factor $2$ cancel out neatly.

Plug in the given data.

You are told $K_\alpha = 4 K_p$ . Also, an alpha particle is a helium nucleus — 2 protons and 2 neutrons — so its mass is approximately 4 times the proton mass: $m_\alpha = 4 m_p$ .

Substitute:

$\frac{\lambda_\alpha}{\lambda_p} = \sqrt{\frac{m_p \cdot K_p}{(4 m_p) \cdot (4 K_p)}} = \sqrt{\frac{1}{16}} = \frac{1}{4}.$ …
Start your 14-day free trial to see the full solution →
CBSE 2025Set X11 markMCQ
$Q. A ball is dropped from a certain height and it falls freely under gravity. During the fall, the de Broglie wavelength associated with it :$
(a) keeps increasing
(b) keeps decreasing
(c) is zero
(d) may increase or decrease
›Reveal solutionSolution

(b) keeps decreasing. The de Broglie wavelength is $\lambda = \dfrac{h}{mv}$ . As the ball falls freely, its speed $v$ increases continuously, so $\lambda$ (inversely proportional to $v$ ) keeps dec …

Start your 14-day free trial to see the full solution →
CBSE 2025Set D1 markMCQ
$Q. The wavelength of de Broglie wave associated with any moving particle does not depend on (A) mass (B) charge (C) velocity (D) momentum$
›Reveal solutionSolution

λ = h/p = h/mv, so it depends on mass, velocity and momentum — charge does not appear in the formula.

The de Broglie wavelength of a moving particle is

λ = h/p = h/(mv)

where h is Planck's constant, m the mass, v the velocity and p = mv the momentum. The formula contains mass, velocity and momentum, so the wavelength depends on all three.

…

Start your 14-day free trial to see the full solution →
CBSE 2025Set A1 mark
$Q. Match Column 'A' item 'Matter waves' with the correct option from Column 'B' and write the correct pair. Column 'B' options:$
(i) Minimum energy to emit electrons from the surface
(ii) Minimum frequency to emit electrons from the surface
(iii) Frequency of photon
(iv) Number of photons
(v) Moving particle
(vi) Photon
(vii) Einstein.
›Reveal solutionSolution

Matter waves correspond to option (v): a moving particle.

Louis de Broglie proposed that, just as light exhibits both wave and particle nature, every moving material particle (electron, proton, or even a macroscopic object) has a wave associated with it, called the matter wave (or de Broglie wave), with wavelength $\lambda = h/p$ , where p is the particle's momentum. This concept applies specifically to a moving particle — a particle at rest (p = 0) has an undef …

Start your 14-day free trial to see the full solution →
CBSE 2025Set ANNUAL1 markMCQ
$Q. Which of the following has the longest de-Broglie wavelength, if they are moving with same velocity ?$
(a) proton
(b) neutron
(c) α-particle
(d) β-particle
›Reveal solutionSolution

At a common velocity, the lightest particle has the longest de Broglie wavelength.

The de Broglie wavelength is

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

For a fixed velocity v (and fixed h), $\lambda \propto 1/m$ — the smaller the mass, the longer the wavelength.

…

Start your 14-day free trial to see the full solution →

←Previous questionQ45 — How do enzymes help a substrate to be attacked by the reagent effectiv…→Next questionQ47 — How will you distinguish 1° and 2° hydroxyl groups present in glucose?…

Q.Describe the term D- and L-configuration used for amino acids with examples.

The Core Idea: Chirality and the Glyceraldehyde Reference

Step-by-Step: How to Assign D or L

Examples

L-Alanine (the natural form)

D-Alanine (found in bacterial cell walls)

L-Cysteine (a special case)

🎓Unlock everything free for 14 days

Unlock everything free for 14 days