Arrange the following particles in ascending order of de Broglie wave…

The correct answer is Option (2) → (C), (D), (B), (A)

The de Broglie wavelength is \( \lambda = h/mv \), so it is inversely proportional to the momentum: the particle with the largest momentum has the shortest wavelength.

    (A) an electron with speed 10⁶ m/s — \( p_A = 9.1\times10^{-31}\,\mathrm{kg} \times 10^{6}\,\mathrm{m\,s^{-1}} = 9.1\times10^{-25}\,\mathrm{kg\,m\,s^{-1}} \)
    (B) a proton with speed of 10⁸ m/s — \( p_B = 1.67\times10^{-27}\,\mathrm{kg} \times 10^{8}\,\mathrm{m\,s^{-1}} = 1.67\times10^{-19}\,\mathrm{kg\,m\,s^{-1}} \)
    (C) a cricket ball of mass 200 g with speed of 10⁶ m/s — \( p_C = 0.2\,\mathrm{kg} \times 10^{6}\,\mathrm{m\,s^{-1}} = 2\times10^{5}\,\mathrm{kg\,m\,s^{-1}} \)
    (D) an alpha particle with speed of 10⁸ m/s — \( p_D = 6.64\times10^{-27}\,\mathrm{kg} \times 10^{8}\,\mathrm{m\,s^{-1}} = 6.64\times10^{-19}\,\mathrm{kg\,m\,s^{-1}} \)

The momenta in descending order are C, D, B, A, so the wavelengths in ascending order are (C), (D), (B), (A).

Shortcut: The alpha particle and the proton have the same speed, and the alpha particle is about four times as heavy, so \( \lambda_D \lt \lambda_B \). The cricket ball is enormously heavier than all the others and the electron is the lightest and slowest, so C comes first and A last.