What will be the de Broglie wavelength of a ball of mass 0.12 kg moving with a speed of 20 m s⁻¹?
- (a)6.63 × 10⁻³⁰ m
- (b)6.63 × 10⁻³⁴ m
- (c)2.76 × 10⁻³⁰ m
- (d)2.76 × 10⁻³⁴ m
Correct — D, 2.76 × 10⁻³⁴ m.
de Broglie's relation assigns a wavelength to a moving body: λ = h/p, where p = mv. For this ball the momentum is 0.12 kg × 20 m s⁻¹ = 2.4 kg m s⁻¹.
So λ = 6.63 × 10⁻³⁴ ÷ 2.4 = 2.76 × 10⁻³⁴ m. Dividing by 2.4 changes only the digits; because 2.4 is a number of order one, the power of ten stays at 10⁻³⁴.
The idea to carry away is that the wavelength is Planck's constant divided by momentum, not Planck's constant by itself. The whole item is one division, and the exponent is decided by the size of mv.
- (a)6.63 × 10⁻³⁰ m — This keeps Planck's constant's digits undivided and also shifts the power of ten by four places, so it fails on both halves of the calculation.
The number is the de Broglie wavelength of a body whose momentum is about 1 × 10⁻⁴ kg m s⁻¹ — for instance a 5 milligram object moving at 20 m s⁻¹, not a 0.12 kg ball.
- (b)6.63 × 10⁻³⁴ m — The digits here are the ones you start from, not the ones you finish with. Picking it means the division by the momentum 2.4 kg m s⁻¹ was never carried out.
Read in the right units, this number is Planck's constant itself, h ≈ 6.63 × 10⁻³⁴ joule-second — a constant of action, and it is measured in J s, not in metres.
- (c)2.76 × 10⁻³⁰ m — The digits are right, so the division 6.63 ÷ 2.4 was done correctly; the power of ten then drifted from 10⁻³⁴ to 10⁻³⁰.
A wavelength of 2.76 × 10⁻³⁰ m needs a momentum of 2.4 × 10⁻⁴ kg m s⁻¹, which is a 12 milligram ball at the same 20 m s⁻¹ — a mass ten thousand times smaller than the one printed in the stem.
Louis de Broglie proposed in 1924 that matter carries a wave character, with wavelength λ = h/p. For a non-relativistic particle p = mv, so λ = h/(mv).
The relation is symmetric with the photon picture: a photon of wavelength λ carries momentum h/λ, and de Broglie turned the statement around to apply to particles with mass.
Because h is about 6.63 × 10⁻³⁴ J s, the wavelength is appreciable only when mv is itself extremely small — the regime of electrons and other light particles.
This single relation is the hinge between classical mechanics and quantum mechanics. It explains why a beam of electrons can be diffracted by a crystal while a cricket ball simply flies in a straight line.
It also underpins Bohr's quantisation condition, since a stable orbit of circumference 2πr holds a whole number of de Broglie wavelengths, and it is the reason electron microscopes resolve far finer detail than light microscopes.
Used numerically, the relation is a single substitution, so the mark rests on keeping the mass in kilograms and tracking the power of ten through the division.
- de Broglie's relation is λ = h/p, with p = mv for a particle moving well below the speed of light.
- Planck's constant h ≈ 6.626 × 10⁻³⁴ J s; the rounded 6.63 × 10⁻³⁴ is what the option set uses.
- For this ball p = 0.12 kg × 20 m s⁻¹ = 2.4 kg m s⁻¹, giving λ = 6.63 × 10⁻³⁴ / 2.4 ≈ 2.76 × 10⁻³⁴ m.
- Louis de Broglie put the matter-wave hypothesis in his 1924 doctoral thesis and received the Nobel Prize in Physics in 1929.
- Davisson and Germer, in 1927, obtained diffraction of electrons from a nickel crystal, supporting the matter-wave idea.
- An electron accelerated through a potential of V volts has a de Broglie wavelength of roughly 1.23/√V nanometres.
- The ball's 10⁻³⁴ m wavelength is far shorter than atomic spacings of about 10⁻¹⁰ m, well below the scale at which diffraction can be detected.
- Since λ = h/(mv), raising either the mass or the speed shortens the de Broglie wavelength in inverse proportion.
One formula, three bodies: the larger the momentum in the denominator, the shorter the matter wave.
- Recalling h = 6.63 × 10⁻³⁴ and marking the option that reproduces it — that is the constant, not the wavelength.
- Mixing units before forming the momentum: h in J s goes with mass in kilograms and speed in m s⁻¹, so a mass quoted in grams must be converted first.
- Getting 6.63 ÷ 2.4 = 2.76 right and then moving the exponent; dividing by 2.4 leaves 10⁻³⁴ untouched.
- Substituting into E = hν, the photon energy–frequency relation, when the wavelength of a massive body needs λ = h/mv.
- Forgetting that momentum, not mass alone, sits in the denominator, so the speed must be multiplied in before dividing.
The relation turns up as a direct substitution, as here, where the option set is built from two mantissas and two exponents so that both the division and the power of ten have to be right.
Read as a pair of choices, the four options make that structure plain: 6.63 or 2.76 for the digits, 10⁻³⁰ or 10⁻³⁴ for the scale, and only 2.76 paired with 10⁻³⁴ survives both steps.
CDS_GK_2022_II_Q242022Same opening step: momentum as p = mv from a stated mass and speed, exactly the quantity that goes into the denominator here. The CDS item stops at that momentum (and its kinetic energy) and stays entirely within classical mechanics, whereas this UKPSC item takes p one step further and divides Planck's constant by it.
NDA_GAT_2025_I_Q1332025Both items ask for a relation involving wavelength, but of different kinds. The NDA question wants v = λf for a sound wave, where wavelength follows from the wave's own speed and frequency; the de Broglie wavelength here comes from Planck's constant divided by the particle's momentum, and involves no propagation speed of a medium.
- practice — not a real PYQ
A stone of mass 0.05 kg moves with a speed of 10 m s⁻¹. Taking h = 6.63 × 10⁻³⁴ J s, its de Broglie wavelength is closest to
- (a)1.33 × 10⁻³³ m
- (b)1.33 × 10⁻³⁴ m
- (c)3.32 × 10⁻³³ m
- (d)6.63 × 10⁻³⁴ m
Answera — momentum is 0.05 × 10 = 0.5 kg m s⁻¹, so λ = 6.63 × 10⁻³⁴ / 0.5 = 1.33 × 10⁻³³ m. Dividing by a number smaller than one pushes the exponent up from −34 to −33.Option b has the right digits but keeps the original exponent. Option c is what you get by dividing by 0.2 rather than 0.5. Option d is Planck's constant reproduced without any division.
- practice — not a real PYQ
If the speed of a moving particle is doubled while its mass stays the same, its de Broglie wavelength (speed well below that of light)
- (a)becomes four times as large
- (b)is halved
- (c)is doubled
- (d)remains unchanged
Answerb — λ = h/(mv), so wavelength is inversely proportional to speed at fixed mass; doubling v halves λ.Option a would require an inverse-square dependence, and option c reverses the proportionality by treating λ as proportional to v. Option d would hold only if λ were independent of motion, whereas the momentum mv sits in the denominator.
- practice — not a real PYQ
The wave nature of electrons, predicted by de Broglie, found experimental support in which one of the following?
- (a)Millikan's oil-drop experiment
- (b)Rutherford's alpha-particle scattering experiment
- (c)Davisson and Germer's electron diffraction experiment
- (d)Stern and Gerlach's silver-atom beam experiment
Answerc — Davisson and Germer scattered electrons from a nickel crystal in 1927 and obtained a diffraction pattern, behaviour characteristic of waves.Millikan's oil-drop work measured the charge on the electron. Rutherford's scattering of alpha particles established the small, massive nucleus. The Stern–Gerlach beam demonstrated space quantisation of magnetic moment, not a wavelength.