Q.A microscope using suitable photons is employed to locate an electron in an atom within a distance of 0.1 Å. What is the uncertainty involved in the measurement of its velocity?

Showing the 12 most recent of 70 on this concept.

1. 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.

   For single-slit diffraction, minima occur at angles $\theta$ satisfying:

   $a \sin\theta = n\lambda$

   where $a$ is the slit width, $n = 1, 2, 3, \ldots$, and $\lambda$ is the de Broglie wavelength.

   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.

   Tip

   A useful way to remember this: small wavelength (high momentum) → more particle-like behavior; large wavelength (low momentum) → more wave-like behavior. But both aspects are always present.

   Connecting Assertion and Reason

   Now the critical question: does the Reason explain the Assertion?

   The answer is yes. The diffraction pattern observed when electrons pass through a slit is a direct consequence of their wave nature. Without wave-like behavior, electrons would simply pass through the slit in straight lines, creating a single bright spot on a screen—no fringes, no pattern.

   The wave-particle duality is not just a parallel fact; it is the cause of the diffraction. The de Broglie wavelength associated with the electron's momentum determines the spacing and angular positions of the diffraction maxima and minima. The Reason provides the fundamental physical principle that makes the Assertion true.

   Watch out

   Don't confuse "both are true" with "one explains the other." Many assertion-reason questions have two correct statements that are unrelated. Here, the causal link is explicit: wave nature → diffraction.

   ✓Final answer

   The correct option is (A): Both Assertion and Reason are true, and Reason is the correct explanation of the Assertion.

2. CBSE 2026Set 55/3/11 markMCQ
   Q.A proton and an alpha particle have equal momentum. The ratio of their kinetic energies $\left(\dfrac{E_p}{E_\alpha}\right)$ and the ratio of the de Broglie wavelengths associated with them $\left(\dfrac{\lambda_p}{\lambda_\alpha}\right)$ respectively are : (A) $2,\ 1$ (B) $1,\ 2$ (C) $4,\ 1$ (D) $1,\ 4$

   ›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.

   1. 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:

   $\frac{\lambda_p}{\lambda_\alpha} = 1$

   2. 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 $m_\alpha$):

   $\frac{E_p}{E_\alpha} = \frac{p^2 / 2m_p}{p^2 / 2m_\alpha} = \frac{m_\alpha}{m_p}$

   3. Know the masses involved. An alpha particle is a helium nucleus — 2 protons and 2 neutrons. Its mass is approximately 4 times the mass of a proton:

   $m_\alpha = 4 m_p$

   This is a standard fact you should remember for such problems.

   Substituting:

   $\frac{E_p}{E_\alpha} = \frac{4 m_p}{m_p} = 4$

   4. Combine the results. We have:

   $\frac{E_p}{E_\alpha} = 4 \quad \text{and} \quad \frac{\lambda_p}{\lambda_\alpha} = 1$

   This matches option (C).

   Watch out

   A common mistake is to forget that kinetic energy depends on mass even when momentum is fixed. Students sometimes assume $E_k \propto p$ and get the ratio wrong. Always write $E_k = p^2/2m$ explicitly.

   Tip

   For any two particles with the same momentum, the de Broglie wavelength ratio is always 1 — it doesn’t matter what the particles are. The kinetic energy ratio is simply the inverse ratio of their masses.

   ✓Final answer

   The correct option is (C), with $\dfrac{E_p}{E_\alpha} = 4$ and $\dfrac{\lambda_p}{\lambda_\alpha} = 1$.

3. CBSE 2026Set V11 markMCQ
   Q.An $\alpha$-particle, a proton, an electron and a neutron are moving with the same velocity. Then the particle having longest de Broglie wavelength is :

   (a) proton

   (b) electron

   (c) neutron

   (d) $\alpha$-particle

   ›Reveal solutionSolution

   (b) electron

   ✓Final answer

   (b) electron

   The de Broglie wavelength is $\lambda = \dfrac{h}{mv}$. For the same velocity $v$, $\lambda \propto \dfrac{1}{m}$, so the particle with the smallest mass has the longest wavelength. Among an $\alpha$-particle, proton, electron and neutron, the electron has by far the smallest mass, so it has the longest de Broglie wavelength.

4. 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

   $\frac{\lambda_1}{\lambda_2} = \sqrt{\frac{V_2}{V_1}} = \sqrt{\frac{64}{49}} = \frac{8}{7}.$

   (The electron through the larger voltage, 64 V, has the shorter wavelength.)

   ✓Final answer

   (D) 8/7.

5. 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 same de Broglie wavelength, even though their masses (and hence their speeds and kinetic energies, for the same $p$) are very different.

   ✓Final answer

   (c) $\lambda_\alpha = \lambda_p = \lambda_e$.

6. 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.

   mv = 0.10 x 10 = 1 kg.m/s

   lambda = 6.63 x 10^-34 / 1 = 6.63 x 10^-34 m

   ✓Final answer

   lambda = 6.63 x 10^-34 m.

7. 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.”

   This is false. Photons do have momentum, given by $p = \frac{h}{\lambda}$ (or $p = \frac{E}{c}$). This momentum is responsible for radiation pressure and is used in technologies like solar sails and optical tweezers. Electrons also have momentum ($p = mv$), but that doesn’t mean photons lack it.

   3. Statement (C): “Photons of light and electrons both exhibit dual nature.”

   This is true. Both show wave-like behaviour (diffraction, interference) and particle-like behaviour (momentum, discrete energy exchange). For photons, the particle aspect is evident in the photoelectric effect; for electrons, the wave aspect is seen in diffraction patterns.

   4. Statement (D): “All electromagnetic radiations do not have photons.”

   This is false. All electromagnetic radiation — from radio waves to gamma rays — consists of photons. The photon is the quantum of the electromagnetic field. There is no EM radiation without photons.

   ---

   ✓Final answer

   The correct option is (C).

8. CBSE 2025Set 55/5/11 markMCQ
   Q.The kinetic energy of an alpha particle is four times the kinetic energy of a proton. The ratio $\dfrac{\lambda_{\alpha}}{\lambda_{p}}$ of the de Broglie wavelengths associated with them will be: (A) $\dfrac{1}{16}$ (B) $\dfrac{1}{8}$ (C) $\dfrac{1}{4}$ (D) $\dfrac{1}{2}$

   ›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.

   1. 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: $\lambda_p = \frac{h}{p_p} = \frac{h}{\sqrt{2 m_p K_p}}$.

   2. 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.

   3. 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:

   $\frac{\lambda_\alpha}{\lambda_p} = \sqrt{\frac{m_p \cdot K_p}{(4 m_p) \cdot (4 K_p)}} = \sqrt{\frac{1}{16}} = \frac{1}{4}.$

   Watch out

   A common mistake is to use $\lambda \propto 1/\sqrt{K}$ without accounting for mass. If you blindly write $\lambda \propto 1/\sqrt{K}$, you’d get $\lambda_\alpha/\lambda_p = 1/2$ — which is wrong. Always include the mass factor from $p = \sqrt{2mK}$.

   Tip

   You can also think in terms of momentum directly: since $K_\alpha = 4K_p$ and $m_\alpha = 4m_p$, we have $p_\alpha = \sqrt{2 \cdot 4m_p \cdot 4K_p} = \sqrt{32 m_p K_p} = 4 \sqrt{2 m_p K_p} = 4 p_p$. So the alpha particle has 4 times the momentum of the proton, and since $\lambda \propto 1/p$, the wavelength ratio is $1/4$.

   ✓Final answer

   The ratio is $\boxed{\frac{1}{4}}$, which corresponds to option (C).

9. 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

   ✓Final answer

   (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 decreasing.

10. 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.

    Electric charge does not appear in the expression, so the de Broglie wavelength does not depend on the charge of the particle.

    ✓Final answer

    (B) charge.

11. 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 undefined (infinite) de Broglie wavelength, so 'matter waves' are intrinsically tied to a moving particle. Hence 'Matter waves' matches '(v) Moving particle'.

    ✓Final answer

    Matter waves → (v) Moving particle.

12. 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.

    Comparing masses: proton ≈ 1836 $m_e$, neutron ≈ 1839 $m_e$, α-particle (2 protons + 2 neutrons) ≈ 4 × 1836 $m_e$, β-particle = a fast electron, mass = $m_e$ (the smallest of all four by a huge margin).

    ✓Final answer

    (d) β-particle — being the least massive, it has the largest de Broglie wavelength at equal velocity.