Q.An electron microscope uses electrons accelerated by a voltage of $50\ \text{kV}$. Determine the de Broglie wavelength associated with the electrons. If other factors (such as numerical aperture, etc.) are taken to be roughly the same, how does the resolving power of an electron microscope compare with that of an optical microscope which uses yellow light?

Showing the 12 most recent of 15 on this concept.

1. COMEDK 2026Set 2026-A1 markMCQ
   Q.The basic principle used behind the working of electron microscope is: (A) Using charged mirrors to achieve the desired magnification. (B) Wave nature of electrons. (C) Electrostatic field created by a beam of electrons. (D) Magnifying power of very thin aperture convex lenses.

   ›Reveal solutionSolution

   The electron microscope relies on the wave nature of electrons to achieve high resolution, far beyond what light microscopes can offer. The correct answer is (B).

   The key insight is that resolution in any microscope is fundamentally limited by the wavelength of the probing radiation. Light microscopes are limited by the wavelength of visible light (about 400–700 nm). To see smaller objects (like viruses or molecules), we need a "probe" with a much shorter wavelength. Electrons, when accelerated, behave as waves (de Broglie wavelength) that can be thousands of times shorter than light waves. This wave nature is what makes electron microscopy possible — not mirrors, electrostatic fields from the beam itself, or thin lenses.

   1. Why not (A) "charged mirrors"?

      Electron microscopes use magnetic coils (electromagnetic lenses) to focus electrons, not charged mirrors. While electrostatic lenses exist, they are not the core principle; the resolution comes from the electron wavelength.

   2. Why not (C) "electrostatic field created by a beam of electrons"?

      The beam's own field is a nuisance (causes beam spreading), not the operating principle. The microscope uses external magnetic fields to shape the beam.

   3. Why not (D) "magnifying power of very thin aperture convex lenses"?

      Convex lenses are for light. Electrons are focused by magnetic fields. Apertures are used to block scattered electrons, but they don't provide magnification.

   4. Why (B) is correct: The wave nature of electrons

      In 1924, Louis de Broglie proposed that particles have a wavelength:

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

   where $h$ is Planck’s constant and $p$ is momentum. For an electron accelerated through a voltage $V$,

   $$$$

   🔒 Start your 14-day free trial to see the full solution →
2. COMEDK 2026Set 2026-M1 markMCQ
   Q.What is the ratio of de Broglie wavelength of an electron to that of proton if the velocity of proton is $\frac{1}{6}$ the velocity of electron? (A) 603:1 (B) 306:1 (C) 1:306 (D) 1:603

   ›Reveal solutionSolution

   The de Broglie wavelength is inversely proportional to momentum. Given the velocity ratio and known masses, the wavelength ratio of electron to proton is 306:1, so the correct option is (B).

   ---

   The key idea is the de Broglie relation:

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

   where $h$ is Planck’s constant, $m$ is mass, and $v$ is velocity.

   Since $h$ is constant, the ratio of wavelengths depends only on the ratio of momenta.

   We are told the proton’s velocity is $\frac{1}{6}$ the electron’s velocity, so we need the mass ratio of proton to electron.

   ---

   1. Write the ratio of wavelengths

   $\frac{\lambda_e}{\lambda_p} = \frac{h/(m_e v_e)}{h/(m_p v_p)} = \frac{m_p v_p}{m_e v_e}$

   2. Insert the given velocity relation $v_p = \frac{1}{6} v_e$, so

   $\frac{\lambda_e}{\lambda_p} = \frac{m_p \cdot \frac{1}{6} v_e}{m_e v_e} = \frac{m_p}{6 m_e}$

   3. Use the known mass ratio The proton mass is about 1836 times the electron mass:

   $m_p \approx 1836 \, m_e$

   Therefore,

   $\frac{\lambda_e}{\lambda_p} = \frac{1836}{6} = 306$

   4. Interpret the ratio …

   🔒 Start your 14-day free trial to see the full solution →
3. KCET 2025Set D-41 markMCQ
   Q.If we consider an electron and a photon of same de-Broglie wavelength, then they will have same (A) Angular momentum (B) Energy (C) Velocity (D) Momentum

   ›Reveal solutionSolution

   $\lambda = h/p$ links wavelength to momentum alone, so equal de Broglie wavelength forces equal momentum — and nothing else.

   Step 1 — The de Broglie relation.

   For any particle (matter or photon), the de Broglie wavelength is

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

   This is a universal relation: it involves Planck's constant and the momentum, and no other property — not mass, not charge, not speed.

   Step 2 — Apply the given condition.

   We are told $\lambda_{e} = \lambda_{ph} = \lambda$. Therefore

   $p_e = \frac{h}{\lambda} = p_{ph}$

   So the momenta are equal. That already identifies the answer, but let us check that the other three quantities are not forced to be equal — this is what makes the question worth asking.

   Step 3 — Energy is different.

   • Photon (massless, moves at $c$): $E_{ph} = pc = \dfrac{hc}{\lambda}$.
   • Electron (non-relativistic, mass $m$): $E_e = \dfrac{p^2}{2m} = \dfrac{h^2}{2m\lambda^2}$.

   These have completely different dependences on $\lambda$ ($1/\lambda$ versus $1/\lambda^2$) and are equal only by accident, not in general. In fact for typical wavelengths $E_{ph} \gg E_e$.

   Step 4 — Velocity is different. …

   🔒 Start your 14-day free trial to see the full solution →
4. COMEDK 2025Set 2025-A1 markMCQ
   Q.Two subatomic particles 1 and 2, with the same kinetic energies have their de- Broglie wavelengths as $\lambda_1 \& \lambda_2$ and masses as 3 m and 6 m respectively. Determine the ratio $\lambda_1: \lambda_2$. (A) $2: 1$ (B) $1: 2$ (C) $\sqrt{2}: 1$ (D) $1: \sqrt{2}$

   ›Reveal solutionSolution

   For particles with equal kinetic energy, de Broglie wavelength is inversely proportional to the square root of mass. Since masses are 3m and 6m, the ratio λ₁:λ₂ = √(6m/3m) = √2:1, so option (C).

   Concept & Intuition

   The de Broglie wavelength of a particle is given by λ = h/p, where h is Planck’s constant and p is momentum. When two particles have the same kinetic energy, their momenta are not equal — heavier particles move slower for the same kinetic energy. Since kinetic energy K = p²/(2m), we have p = √(2mK). So λ = h/√(2mK). With K constant, λ ∝ 1/√m. That’s the key: wavelength is inversely proportional to the square root of mass.

   Step-by-step reasoning

   1. Write the de Broglie relation

      λ = h / p, where p is momentum.

   2. Express momentum in terms of kinetic energy

      Kinetic energy K = p²/(2m) ⇒ p = √(2mK).

      So λ = h / √(2mK).

   3. Apply to both particles

      For particle 1 (mass m₁ = 3m): λ₁ = h / √(2·3m·K).

      For particle 2 (mass m₂ = 6m): λ₂ = h / √(2·6m·K).

   4. Take the ratio …

   🔒 Start your 14-day free trial to see the full solution →
5. COMEDK 2025Set 2025-A1 markMCQ
   Q.de Broglie wave length associated with an electron accelerated through a potential difference ' V ' is ' $\lambda$ '. If the accelerating potential is halved, what will be the new wave length associated with the charged particle? (A) $\frac{\lambda}{\sqrt{2}}$ (B) $\frac{\lambda}{2}$ (C) $2 \lambda$ (D) $\sqrt{2} \boldsymbol{\lambda}$

   ›Reveal solutionSolution

   The de Broglie wavelength of an electron accelerated through a potential $V$ is inversely proportional to the square root of $V$. Halving $V$ multiplies the wavelength by $\sqrt{2}$, so the new wavelength is $\sqrt{2}\,\lambda$. The correct option is (D).

   The key idea here is the de Broglie relation for a charged particle accelerated from rest. When an electron is accelerated through a potential difference $V$, it gains kinetic energy equal to $eV$. This kinetic energy determines its momentum, and the de Broglie wavelength is inversely proportional to momentum. Since momentum depends on the square root of kinetic energy, it depends on the square root of $V$. Halving $V$ therefore changes the wavelength by a factor of $\sqrt{2}$.

   Let’s work through it step by step.

   1. Relate kinetic energy to accelerating potential An electron of charge $e$ accelerated from rest through a potential difference $V$ gains kinetic energy:

   $K = eV$

   This is the energy it has when it reaches the target.

   2. Express momentum in terms of kinetic energy For a non-relativistic electron (which is valid for typical accelerating potentials up to a few thousand volts), the kinetic energy is related to momentum $p$ by:

   $K = \frac{p^2}{2m}$

   where $m$ is the electron’s mass. Solving for $p$:

   $p = \sqrt{2mK} = \sqrt{2m eV}$

   3. Apply the de Broglie relation The de Broglie wavelength $\lambda$ is:

   $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2m e V}}$

   Here $h$ is Planck’s constant. Notice that $\lambda$ is inversely proportional to $\sqrt{V}$.

   4. Halve the accelerating potential If the new potential is $V' = \frac{V}{2}$, then the new wavelength $\lambda'$ is:

   $\lambda' = \frac{h}{\sqrt{2m e \cdot (V/2)}} = \frac{h}{\sqrt{2m e V \cdot \frac{1}{2}}} = \frac{h}{\sqrt{2m e V} \cdot \frac{1}{\sqrt{2}}}$

   Simplify: …

   🔒 Start your 14-day free trial to see the full solution →
6. COMEDK 2025Set 2025-M1 markMCQ
   Q.A particle of mass $M$ at rest decays into masses $m_1$ and $m_2$ with non-zero velocities. The ratio of de Broglie wavelengths $\lambda_1$ and $\lambda_2$ of the particles is (A) $\frac{m_2}{m_1}$ (B) $\frac{\sqrt{m_1}}{\sqrt{m_2}}$ (C) $\frac{m_1}{m_2}$ (D) $1: 1$

   ›Reveal solutionSolution

   The de Broglie wavelength depends only on momentum, and in a decay from rest, momentum is conserved so the two particles have equal and opposite momenta. Therefore their wavelengths are equal, giving ratio 1:1.

   The key concept here is conservation of momentum combined with the de Broglie relation.

   A common mistake is to think wavelength depends on mass or energy, but the de Broglie wavelength is $\lambda = h/p$, where $p$ is momentum. Since the parent particle is at rest, its total momentum is zero. After decay, the two daughter particles must have equal and opposite momenta to conserve momentum. Thus their momenta are equal in magnitude, and so their de Broglie wavelengths are equal — regardless of their masses.

   1. Recall the de Broglie relation

      For any particle, $\lambda = \frac{h}{p}$, where $h$ is Planck’s constant and $p$ is the magnitude of its momentum.

   2. Apply conservation of momentum

      The parent particle is initially at rest, so total momentum is zero. After decay,

   $\vec{p}_1 + \vec{p}_2 = 0 \quad \Rightarrow \quad \vec{p}_1 = -\vec{p}_2.$

   Hence the magnitudes are equal: $p_1 = p_2$.

   3. Compare wavelengths Since $\lambda_1 = h/p_1$ and $\lambda_2 = h/p_2$, and $p_1 = p_2$, we get $$$$

   🔒 Start your 14-day free trial to see the full solution →
7. COMEDK 2024Set 2024-A1 markMCQ
   Q.The velocity of an electron so that its momentum is equal to that of a photon of wavelength $660 \mathrm{~nm}$ is (A) $109.88 \mathrm{~cm} \mathrm{~s}^{-1}$ (B) $1098 \mathrm{~ms} ^{-1}$ (C) $1098 \mathrm{~cm} \mathrm{~s}^{-1}$ (D) $109 \mathrm{~ms} ^{-1}$

   ›Reveal solutionSolution

   Setting the electron momentum equal to the photon momentum $h/\lambda$ gives an electron speed of about $1098\ \text{m s}^{-1}$.

   Photon momentum:

   $p=\frac{h}{\lambda}=\frac{6.626\times10^{-34}}{660\times10^{-9}}=1.004\times10^{-27}\ \text{kg m s}^{-1}.$

   Electron speed ($p=m_e v$): …

   🔒 Start your 14-day free trial to see the full solution →
8. COMEDK 2024Set 2024-E1 markMCQ
   Q.A particle of mass $2 \mathrm{mg}$ has the same wavelength as a neutron moving with a velocity of $3 \times 10^5 \mathrm{~ms}^{-1}$. The velocity of the particle is (mass of neutron is $1.67 \times 10^{-27} \mathrm{Kg}$) (A) $2.5 \times 10^{-16} \mathrm{~ms}^{-1}$ (B) $1.5 \times 10^{-13} \mathrm{~ms}^{-1}$ (C) $2.5 \times 10^{-13} \mathrm{~ms}^{-1}$ (D) $1.5 \times 10^{-16} \mathrm{~ms}^{-1}$

   ›Reveal solutionSolution

   Same $\lambda\Rightarrow$ same $mv$; the $2\,mg$ particle moves at $\approx2.5\times10^{-16}$ m/s.

   de Broglie: $\lambda=\dfrac{h}{mv}$. Equal wavelengths give equal momenta:

   $m_{p}v_{p}=m_{n}v_{n}$

   Mass of particle $=2\text{ mg}=2\times10^{-6}$ kg. …

   🔒 Start your 14-day free trial to see the full solution →
9. COMEDK 2024Set 2024-M1 markMCQ
   Q.Given below are 4 subatomic particles travelling with the same velocity "v". Which of them will have the shortest wavelength? $[A]=\text { Proton }[B]=\text { Neutron. }[C]=\text { Alpha particle. }[D]=\text { Electron. }$ (A) [B] (B) [D] (C) [C] (D) [A]

   ›Reveal solutionSolution

   At a fixed velocity the de Broglie wavelength is inversely proportional to mass. The alpha particle has the largest mass, so it has the shortest wavelength — the correct option is (C).

   Concept

   Every moving particle has an associated de Broglie wavelength

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

   where $h$ is Planck's constant and $p$ the momentum. Because all four particles move with the same velocity $v$, each momentum is $p = mv$, so

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

   With $h$ and $v$ common to all, the wavelength is inversely proportional to the mass: the heavier the particle, the shorter its wavelength.

   Solution

   Compare the masses:

   1. Electron — $\approx 9.11\times10^{-31}\,\text{kg}$; lightest, so the longest wavelength.
   2. Proton — about 1836 times the electron mass.
   3. Neutron — essentially the same mass as a proton.
   4. Alpha particle — two protons and two neutrons, roughly four proton masses; the heaviest of the four. …

   🔒 Start your 14-day free trial to see the full solution →
10. COMEDK 2024Set 2024-M1 markMCQ
    Q.The mass of a particle $\mathrm{A}$ is double that of the particle $\mathrm{B}$ and the kinetic energy of $\mathrm{B}$ is $\frac{1}{8}$th that of A then the ratio of the de- Broglie wavelength of A to that of B is: (A) 1 : 2 (B) 2 : 1 (C) 1 : 4 (D) 4 : 1

    ›Reveal solutionSolution

    The de Broglie wavelength depends on momentum, which we get from kinetic energy and mass. Using the given ratios, the wavelength ratio is 1:4, so option (C).

    The de Broglie wavelength of a particle is given by $\lambda = \frac{h}{p}$, where $h$ is Planck’s constant and $p$ is the momentum. Since kinetic energy $K = \frac{p^2}{2m}$, we can express momentum as $p = \sqrt{2mK}$. Therefore, the wavelength becomes $\lambda = \frac{h}{\sqrt{2mK}}$. The ratio of wavelengths for two particles depends only on their masses and kinetic energies.

    Let’s denote:

    • Mass of A: $m_A = 2m$ (since A is double B)
    • Mass of B: $m_B = m$
    • Kinetic energy of B: $K_B = \frac{1}{8} K_A$ (since B’s KE is 1/8th of A’s)

    We want $\frac{\lambda_A}{\lambda_B}$.

    1. Write the wavelength formula for each particle

    $\lambda_A = \frac{h}{\sqrt{2 m_A K_A}}, \quad \lambda_B = \frac{h}{\sqrt{2 m_B K_B}}$

    2. Take the ratio

    $\frac{\lambda_A}{\lambda_B} = \frac{h / \sqrt{2 m_A K_A}}{h / \sqrt{2 m_B K_B}} = \sqrt{\frac{m_B K_B}{m_A K_A}}$

    3. Substitute the given ratios $m_A = 2m$, $m_B = m$, and $K_B = \frac{1}{8} K_A$. $$$$

    🔒 Start your 14-day free trial to see the full solution →
11. COMEDK 2023Set 2023-E1 markMCQ
    Q.A particle at rest decays in to two particles of mass $m_1$ and $m_2$ and move with velocities $v_1$ and $v_2$. The ratio of their de Broglie wave length $\frac{\lambda_1}{\lambda_2}$ is: (A) 1 : 4 (B) 1 : 1 (C) 1 : 2 (D) 2 : 1

    ›Reveal solutionSolution

    So lambda1 : lambda2 = 1 : 1, whatever the masses and speeds individually are.

    Concept: conservation of linear momentum in a decay from rest, plus the de Broglie relation lambda = h/p.

    The parent particle is at rest, so total momentum before = 0. After the decay,

    m1 v1 + m2 v2 = 0 => the two fragments fly apart with momenta EQUAL IN MAGNITUDE (and opposite in direction):

    |p1| = |p2| = p.

    de Broglie wavelength: lambda = h / p. …

    🔒 Start your 14-day free trial to see the full solution →
12. COMEDK 2022Set 20221 markMCQ
    Q.Ultraviolet light of wavelength 99 mm falls on a metal plate of work function 1.0 eV. If the mass of the electron is 9.1 $\times$ 10$^{-31}$ kg, the wavelength of the fastest photoelectron emitted is (A) 0.63 nm (B) 0.66 nm (C) 0.33 nm (D) 0.36 nm

    ›Reveal solutionSolution

    (Equivalently lambda = h/sqrt(2mK): K = 11.53 x 1.6e-19 = 1.845e-18 J; sqrt(2 x 9.1e-31 x 1.845e-18) = 1.833e-24 kg m/s; lambda = 6.626e-34/1.833e-24 = 3.6e-10 m.)

    Concept: Einstein's photoelectric equation gives the maximum kinetic energy, then the de Broglie relation gives the wavelength of the fastest electron. (The '99 mm' is a misprint for 99 nm - only a nanometre-scale UV wavelength makes sense.)

    Photon energy:

    E = 1240 eV nm / 99 nm = 12.53 eV

    Maximum kinetic energy:

    K = E - phi = 12.53 - 1.0 = 11.53 eV

    de Broglie wavelength of an electron of kinetic energy K (in eV):

    lambda = 1.226 / sqrt(K) nm …

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