Two particles \( A_1 \) and \( A_2 \) of masses \( m_1 \) and \( m_2 \) \( (m_1\gt m_2) \) have the same de Broglie wavelength. Then
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → the energy of \( A_1 \) is less than the energy of \( A_2 \)
By the de Broglie relation \( \lambda=h/p \), particles with the same wavelength have the same momentum \( p \). Their kinetic energy, written in terms of momentum, is:
\( \begin{aligned} K &= \frac{1}{2}mv^{2} \\ &= \frac{p^{2}}{2m} \end{aligned} \)
For the same \( p \), the kinetic energy is inversely proportional to the mass.
Option (1) their speeds are the same — Incorrect. Equal momenta give \( m_1v_1=m_2v_2 \). With \( m_1\gt m_2 \), the heavier particle moves slower, \( v_1\lt v_2 \).
Option (2) their energies are the same — Incorrect. With the same momentum, \( K=p^{2}/2m \) is different for different masses.
Option (3) the energy of \( A_1 \) is less than the energy of \( A_2 \) — Correct. Both have the same \( p \), and \( A_1 \) has the larger mass, so \( K_1=p^{2}/2m_1 \) is smaller than \( K_2=p^{2}/2m_2 \).
Option (4) the energy of \( A_2 \) is less than the energy of \( A_1 \) — Incorrect. \( A_2 \) is the lighter particle, so with the same momentum it has the larger energy.
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”
\( \displaystyle \lambda=\frac{h}{p} \)\( \displaystyle =\frac{h}{mv} \qquad (11.5) \)