Q.Relativistic corrections become necessary when the expression for the kinetic energy $\frac{1}{2}mv^2$ becomes comparable with $mc^2$, where $m$ is the mass of the particle. At what de Broglie wavelength will relativistic corrections become important for an electron?

Showing the 12 most recent of 42 on this concept.

1. AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
   Q.A proton and an α-particle are accelerated through the same potential difference. The ratio of their de Broglie wavelengths will be (A) 1:1 (B) 1:2 (C) 2:1 (D) $2\sqrt{2}:1$

   ›Reveal solutionSolution

   This tests the de Broglie wavelength of a charged particle accelerated through a potential difference, $\lambda=h/\sqrt{2mqV}$, and how it scales with mass and charge. Answer: $2\sqrt2:1$.

   Concept and Intuition

   When a charge $q$ is accelerated from rest through potential difference $V$, it gains kinetic energy $qV=\tfrac12mv^2$, so its momentum is $p=\sqrt{2mqV}$. Heavier or more highly charged particles gain more momentum for the same accelerating voltage, and since $\lambda=h/p$, they end up with a shorter de Broglie wavelength. Comparing a proton and an alpha particle (which is heavier by 4× and carries twice the charge), the alpha particle picks up momentum that the proton cannot match, so its wavelength is much shorter.

   Step-by-Step Solution

   1. $\lambda=\dfrac{h}{p}=\dfrac{h}{\sqrt{2mqV}}$ for a particle of mass $m$, charge $q$, accelerated through $V$.
   2. For the proton: $m=m_p$, $q=e$ $\Rightarrow \lambda_p=\dfrac{h}{\sqrt{2m_p eV}}$.
   3. For the alpha particle: $m=4m_p$, $q=2e$ $\Rightarrow \lambda_\alpha=\dfrac{h}{\sqrt{2(4m_p)(2e)V}}=\dfrac{h}{\sqrt{16\,m_p eV}}$. …

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2. AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
   Q.If two particles have same kinetic energy, then the ratio of their de Broglie wavelengths ($\lambda_A:\lambda_B$) is (Given: $m_A = 4m_B$) (A) $1:2$ (B) $2:1$ (C) $1:4$ (D) $4:1$

   ›Reveal solutionSolution

   This tests how the de Broglie wavelength depends on mass at fixed kinetic energy: $\lambda\propto1/\sqrt m$. Answer: $\lambda_A:\lambda_B=1:2$.

   Concept and Intuition

   Kinetic energy relates to momentum as $K=\dfrac{p^2}{2m}$, so $p=\sqrt{2mK}$. For a fixed kinetic energy, a heavier particle has larger momentum (it takes more momentum to carry the same energy when mass is larger). Since the de Broglie wavelength $\lambda=h/p$ is inversely proportional to momentum, the heavier particle ends up with the shorter wavelength.

   Step-by-Step Solution

   1. $\lambda=\dfrac hp=\dfrac h{\sqrt{2mK}}$, and since $K$ is the same for both particles, $\lambda\propto\dfrac1{\sqrt m}$.
   2. $\dfrac{\lambda_A}{\lambda_B}=\sqrt{\dfrac{m_B}{m_A}}$.
   3. Given $m_A=4m_B$, so $\dfrac{m_B}{m_A}=\dfrac14$. …

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3. AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
   Q.The de Broglie wavelength associated with a pebble of mass 6.63 $g$ travelling with velocity $x\ ms^{-1}$ is $10^{-24}$ nm. What is the value of $x$? ($h = 6.63 \times 10^{-34}\ Js$) (A) $1.0 \times 10^{-1}$ (B) $1.0 \times 10^{2}$ (C) $1.0 \times 10^{3}$ (D) $1.0 \times 10^{-2}$

   ›Reveal solutionSolution

   This is a direct application of the de Broglie wavelength formula, solved for velocity; $x = 1.0\times10^{2}\ ms^{-1}$.

   Concept and Intuition

   Every moving particle has an associated wavelength $\lambda = h/p = h/(mv)$. For everyday (macroscopic) masses this wavelength is absurdly small — which is exactly why matter waves aren't noticed for pebbles, only for particles like electrons. Careful unit bookkeeping (grams to kg, nm to m) is the whole trick here.

   Step-by-Step Solution

   1. de Broglie relation: $\lambda = \dfrac{h}{mv} \Rightarrow v = \dfrac{h}{m\lambda}$.
   2. Convert mass: $m = 6.63\ g = 6.63\times10^{-3}$ kg.
   3. Convert wavelength: $\lambda = 10^{-24}\ nm = 10^{-24}\times10^{-9}\ m = 10^{-33}$ m. …

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4. AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQ
   Q.If the radius of the first Bohr orbit of $He^+$ is 26.45 $pm$, then the de Broglie wavelength (in $m$) associated with the electron present in its fourth orbit is ($\pi = 3.14$) (A) $3.13 \times 10^{-11}$ (B) $1.33 \times 10^{-11}$ (C) $2.33 \times 10^{-10}$ (D) $3.32 \times 10^{-10}$

   ›Reveal solutionSolution

   Using the Bohr–de Broglie relation, the electron in the fourth orbit of $He^+$ has $\lambda = 2\pi r_1 n$; the official key marks option (D), $3.32\times10^{-10}\,\text{m}$.

   Standing-wave (Bohr–de Broglie) condition. For a stable orbit the circumference holds a whole number of de Broglie wavelengths:

   $n\lambda = 2\pi r_n \implies \lambda = \frac{2\pi r_n}{n}.$

   Scale the radius. For $He^+$, $r_n = r_1 n^2$ with $r_1 = 26.45\ \text{pm}$, so

   $\lambda = \frac{2\pi (r_1 n^2)}{n} = 2\pi r_1 n = 2(3.14)(26.45\ \text{pm})(4) = 3.32\times10^{-10}\,\text{m (as marked)}.$ …

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5. AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
   Q.The radius of first Bohr orbit of hydrogen is $a_0\ nm$. What is the de Broglie wavelength (in $nm$) associated with electron present in fourth orbit of hydrogen? (A) $2\pi a_0$ (B) $\dfrac{2\pi a_0}{3}$ (C) $8\pi a_0$ (D) $4\pi a_0$

   ›Reveal solutionSolution

   This tests Bohr's quantization condition combined with de Broglie's hypothesis: the orbit circumference must be an integral number of de Broglie wavelengths. For $n=4$ in hydrogen, $\lambda = 8\pi a_0$.

   Concept and Intuition

   Bohr postulated angular momentum quantization $mvr_n = \dfrac{nh}{2\pi}$. De Broglie's insight was that this is equivalent to saying the electron's wave must close on itself around the orbit: $2\pi r_n = n\lambda$, i.e., the orbit circumference contains a whole number of wavelengths — this is what makes a stationary (non-radiating) orbit possible.

   Step-by-Step Solution

   1. De Broglie condition: $\lambda = \dfrac{2\pi r_n}{n}$.
   2. Bohr radius formula for hydrogen: $r_n = n^2 a_0$, where $a_0$ is the first Bohr radius (given as $a_0$ nm here). …

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6. AP EAPCET 2025Set eng-2025-05-21-FN1 markMCQ
   Q.If the linear momentum of a proton is changed by $p_0$, then the de Broglie wavelength associated with the proton changes by 0.25%. Then the initial linear momentum of the proton is (A) $100p_0$ (B) $\dfrac{p_0}{400}$ (C) $400p_0$ (D) $\dfrac{p_0}{100}$

   ›Reveal solutionSolution

   This tests the inverse relationship between de Broglie wavelength and momentum, applied to small fractional changes. Answer: $400p_0$.

   Concept and Intuition

   The de Broglie wavelength is $\lambda=\dfrac{h}{p}$ — inversely proportional to momentum. For a small change in momentum, the fractional change in wavelength has the same magnitude as the fractional change in momentum (since $h$ is constant): $\left|\dfrac{\Delta\lambda}{\lambda}\right|=\left|\dfrac{\Delta p}{p}\right|$. This lets us relate a small known change in momentum directly to the resulting fractional change in wavelength, without needing $h$ at all.

   Step-by-Step Solution

   1. $\lambda=\dfrac{h}{p} \Rightarrow \ln\lambda=\ln h-\ln p$.
   2. Differentiating: $\dfrac{\Delta\lambda}{\lambda}=-\dfrac{\Delta p}{p}$, so in magnitude $\dfrac{|\Delta\lambda|}{\lambda}=\dfrac{\Delta p}{p}$. …

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7. AP EAPCET 2025Set eng-2025-05-21-AN1 markMCQ
   Q.A particle of mass $8\ \mu\mathrm{g}$ in motion collides with another stationary particle of mass $4\ \mu\mathrm{g}$. If the collision is perfectly elastic and one dimensional, the ratio of their de Broglie wavelengths after collision is (A) 4 : 1 (B) 3 : 1 (C) 1 : 1 (D) 2 : 1

   ›Reveal solutionSolution

   Using elastic-collision velocity formulas and the de Broglie relation $\lambda=h/p$, the ratio of wavelengths after collision works out to $2:1$.

   Concept and Intuition

   In a 1-D elastic collision between a moving mass $m_1$ and a stationary mass $m_2$, both momentum and kinetic energy are conserved, giving standard formulas for the final velocities. The de Broglie wavelength of each particle depends on its momentum as $\lambda = h/p$, so the wavelength ratio after collision is the inverse ratio of their momenta.

   Step-by-Step Solution

   1. Let $m_1=8\ \mu\text{g}$ (moving with initial speed $u_1$), $m_2=4\ \mu\text{g}$ (initially at rest).
   2. Elastic collision formulas:

   $v_1' = \frac{m_1-m_2}{m_1+m_2}u_1 = \frac{8-4}{12}u_1 = \frac{u_1}{3}$

   $v_2' = \frac{2m_1}{m_1+m_2}u_1 = \frac{16}{12}u_1 = \frac{4u_1}{3}$

   3. Post-collision momenta: $p_1 = m_1v_1' = 8\times\frac{u_1}{3} = \frac{8u_1}{3}$; $p_2 = m_2v_2' = 4\times\frac{4u_1}{3} = \frac{16u_1}{3}$. …

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8. AP EAPCET 2025Set eng-2025-05-22-FN1 markMCQ
   Q.An alpha particle moves along a circular path of radius 0.5 mm in a magnetic field of $2 \times 10^{-2}$ T. The de Broglie wavelength associated with the alpha particle is nearly (Planck's constant = $6.63 \times 10^{-34}$ J s) (A) 3.1 Å (B) 1.1 Å (C) 0.1 Å (D) 2.1 Å

   ›Reveal solutionSolution

   The circular radius fixes the alpha particle's momentum via $p=qBr$; the de Broglie wavelength $\lambda=h/p$ then comes out to about 2.1 Å.

   Concept and Intuition

   A charged particle moving perpendicular to a uniform magnetic field travels in a circle because the magnetic force provides exactly the centripetal force needed: $qvB = \dfrac{mv^2}{r}$, which rearranges to $mv = qBr$, i.e. the particle's momentum $p=mv$ is directly proportional to $B$ and $r$. This is a very useful shortcut — it lets us find momentum from the geometry of the path (radius) and the field, without separately knowing mass and velocity. Once we have the momentum, the de Broglie relation $\lambda = h/p$ (matter behaving as a wave, wavelength inversely proportional to momentum) gives the wavelength directly.

   Step-by-Step Solution

   1. From the circular-motion condition: $qvB = \dfrac{mv^2}{r} \Rightarrow mv = p = qBr$.
   2. Charge of an alpha particle: $q = 2e = 2\times1.6\times10^{-19} = 3.2\times10^{-19}$ C.
   3. Substitute $B=2\times10^{-2}$ T, $r=0.5\ \mathrm{mm}=5\times10^{-4}$ m: $p = (3.2\times10^{-19})(2\times10^{-2})(5\times10^{-4}) = 3.2\times10^{-24}\ \mathrm{kg\,m/s}.$ …

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9. AP EAPCET 2025Set eng-2025-05-23-FN1 markMCQ
   Q.If a proton and an alpha particle are accelerated through the same potential difference, then the ratio of their de Broglie wavelengths is (A) 1 : 2 (B) 1 : 4 (C) $2\sqrt{2} : 1$ (D) 1 : 8

   ›Reveal solutionSolution

   A charged particle accelerated through potential $V$ gains kinetic energy $qV$, giving momentum $p=\sqrt{2mqV}$; the de Broglie wavelength ratio then depends on both the mass ratio and the charge ratio of the two particles.

   Concept and Intuition

   When a particle of charge $q$ and mass $m$ is accelerated from rest through potential difference $V$, it gains kinetic energy $qV = \tfrac12 mv^2$, so its momentum is $p = mv = \sqrt{2mqV}$. The de Broglie wavelength is $\lambda = h/p = h/\sqrt{2mqV}$. Because both mass and charge differ between a proton and an alpha particle, both ratios must be tracked — it's a common trap to only account for the mass difference.

   Step-by-Step Solution

   1. Proton: mass $m_p$, charge $q_p = e$.
   2. Alpha particle: mass $m_\alpha = 4m_p$ (2 protons + 2 neutrons), charge $q_\alpha = 2e$.
   3. $\lambda_p = \dfrac{h}{\sqrt{2m_p q_p V}}$, $\lambda_\alpha = \dfrac{h}{\sqrt{2m_\alpha q_\alpha V}}$. …

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10. AP EAPCET 2025Set eng-2025-05-23-FN1 markMCQ
    Q.The deBroglie wavelengths of two fast moving particles X, Y are 1 nm, 3 nm respectively. Mass of X is nine times the mass of Y. The ratio of kinetic energies of X, Y is (A) $1:3$ (B) $1:1$ (C) $9:1$ (D) $1:9$

    ›Reveal solutionSolution

    This tests combining the de Broglie relation with the kinetic-energy–momentum relation; the ratio comes out 1:1, option (B).

    Concept and Intuition

    The de Broglie wavelength is $\lambda = \dfrac{h}{p}$, so momentum $p = \dfrac{h}{\lambda}$. Kinetic energy in terms of momentum is $KE = \dfrac{p^2}{2m}$. Substituting gives $KE = \dfrac{h^2}{2m\lambda^2}$ — so KE depends on both mass and wavelength, and a heavier particle with a proportionally longer wavelength can end up with the same kinetic energy as a lighter, shorter-wavelength one.

    Step-by-Step Solution

    1. Write $KE_X = \dfrac{h^2}{2m_X\lambda_X^2}$ and $KE_Y = \dfrac{h^2}{2m_Y\lambda_Y^2}$.
    2. Take the ratio: $\frac{KE_X}{KE_Y} = \frac{m_Y \lambda_Y^2}{m_X \lambda_X^2}$ …

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11. AP EAPCET 2025Set ap-2025-05-19-FN1 markMCQ
    Q.A particle is moving with a velocity four times the velocity of an electron. If the de Broglie wavelength of the particle is $1.5\times10^{-4}$ times the de Broglie wavelength of the electron, then the mass of the particle is (mass of the electron $= 9\times10^{-31}\ \mathrm{kg}$) (A) $1.5\times10^{-31}$ kg (B) $1.5\times10^{-27}$ kg (C) $2.25\times10^{-27}$ kg (D) $2.25\times10^{-31}$ kg

    ›Reveal solutionSolution

    Using $\lambda = h/(mv)$ and the ratio of wavelengths and speeds given, the unknown particle's mass comes out to $1.5\times10^{-27}\,\mathrm{kg}$.

    Concept and Intuition

    The de Broglie wavelength of a particle is $\lambda = \dfrac{h}{mv}$ — inversely proportional to the momentum $mv$. Comparing two particles, the ratio of their wavelengths equals the inverse ratio of their momenta, letting us solve for an unknown mass given the speed and wavelength ratios.

    Step-by-Step Solution

    1. $\lambda_p = \dfrac{h}{m_p v_p}$, $\lambda_e = \dfrac{h}{m_e v_e}$, so $\dfrac{\lambda_p}{\lambda_e} = \dfrac{m_e v_e}{m_p v_p}$.
    2. Given $v_p = 4v_e$, so $\dfrac{v_e}{v_p} = \dfrac14$.
    3. Given $\dfrac{\lambda_p}{\lambda_e} = 1.5\times10^{-4}$.
    4. So $1.5\times10^{-4} = \dfrac{m_e}{m_p}\times\dfrac14 \Rightarrow \dfrac{m_e}{m_p} = 6\times10^{-4}$. …

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12. AP EAPCET 2025Set ap-2025-05-20-AN1 markMCQ
    Q.If a bullet of mass 2.2 g travels with a velocity of $3 \times 10^4\,\mathrm{cm\,s^{-1}}$, its wavelength (in m) is $(h = 6.6 \times 10^{-34}\,\mathrm{Js})$ (A) $10^{-36}$ (B) $10^{-31}$ (C) $10^{-34}$ (D) $10^{-33}$

    ›Reveal solutionSolution

    A direct application of the de Broglie wavelength formula after careful unit conversion, giving $10^{-33}$ m.

    Concept and Intuition

    Every moving particle has an associated de Broglie wavelength $\lambda=h/p$. For macroscopic objects like a bullet, this wavelength is astronomically small, which is why matter waves are unobservable at everyday scales.

    Step-by-Step Solution

    1. Convert mass: $2.2\ \text{g} = 2.2\times10^{-3}$ kg.
    2. Convert velocity: $3\times10^4\ \text{cm s}^{-1} = 3\times10^2\ \text{m s}^{-1}=300$ m/s.
    3. Momentum: $p=mv=2.2\times10^{-3}\times300=0.66$ kg·m/s. …

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