The de Broglie wavelength is given by \( \lambda=h/p \), where p is the momentum of the particle. If the following particles move with the same speed, for which will the de Broglie wavelength be the least?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → alpha particle
By the de Broglie relation, \( \lambda=h/p \) with momentum \( p=mv \). For particles moving with the same speed, \( \lambda\propto1/m \), so the heaviest particle has the shortest wavelength.
Option (1) electron — Incorrect. The electron (\( m\approx9.11\times10^{-31}\,\mathrm{kg} \)) is the lightest of the four, about 1/1836 of the proton mass. So at the same speed it has the smallest momentum and the longest wavelength.
Option (2) proton — Incorrect. The proton (\( m_p\approx1.67\times10^{-27}\,\mathrm{kg} \)) has about one-fourth the mass of an alpha particle, so its wavelength is about four times longer.
Option (3) neutron — Incorrect. The neutron mass is almost equal to the proton mass. It is again about one-fourth of the alpha particle’s mass, so its wavelength is also about four times longer.
Option (4) alpha particle — Correct. An alpha particle has two protons and two neutrons, so \( m_\alpha\approx4m_p \). It is the heaviest of the four, has the largest momentum \( mv \) at the same speed and hence the shortest de Broglie wavelength.
NCERT (2026-27):
Wave Nature of Matter – “… De Broglie proposed that the wave length λ associated with a particle of momentum p is given as
\( \displaystyle \lambda=\frac{h}{p}=\frac{h}{mv} \)\( \qquad (11.5) \)
where m is the mass of the particle and v its speed. … λ is smaller for a heavier particle (large m) or more energetic particle (large v).”