The following particles have the same kinetic energy. Arrange them in…

The correct answer is Option (1) → (C), (B), (A), (D)

For a particle of kinetic energy K, the momentum is \( p = \sqrt{2mK} \), so the de Broglie wavelength is

    \( \displaystyle \lambda = \frac{h}{\sqrt{2mK}} \)

With the same K, \( \lambda \propto 1/\sqrt{m} \): the heavier the particle, the shorter its wavelength. The masses are:

    \( m_\alpha \approx 4\,\mathrm{u} \)
    \( m_n = 1.00866\,\mathrm{u} \)
    \( m_p = 1.00728\,\mathrm{u} \)
    \( m_\beta = m_e \approx 0.00055\,\mathrm{u} \)

So \( m_\alpha \gt m_n \gt m_p \gt m_e \), and the wavelengths increase in the order α-particle, neutron, proton, β-particle: (C), (B), (A), (D).