Following particles are having the same kinetic energy. Arrange them in the ascending order of their de-Broglie wavelengths.
(A) An electron
(B) A proton
(C) A neutron
(D) An α-particle
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (D), (C), (B), (A)
The de Broglie wavelength is \( \lambda = h/p \), and for a particle of kinetic energy K the momentum is \( p = \sqrt{2mK} \).
\( \displaystyle \lambda = \frac{h}{\sqrt{2mK}} \)
With the same K for all four particles, \( \lambda \propto 1/\sqrt{m} \): the heavier the particle, the shorter its wavelength. Their masses are in the order
\( m_\alpha \approx 4m_p \gt m_n \gt m_p \gt m_e \)
(the neutron is slightly heavier than the proton, and the electron is about 1836 times lighter than the proton).
Ascending order of wavelength: (D), (C), (B), (A).