Arrange the following particles having same kinetic energy in increasing order of their de-Broglie wavelengths.
A. Electron
B. Proton
C. Neutron
D. Alpha particle
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → D, C, B, A
For a particle of mass m and kinetic energy K, the de Broglie wavelength is \( \lambda = h/p = h/\sqrt{2mK} \). With the same K, \( \lambda \propto 1/\sqrt{m} \): the heavier the particle, the shorter its wavelength.
\( \text{(D) Alpha particle: } m \approx 4.0015\,\mathrm{u} \)
\( \text{(C) Neutron: } m = 1.00866\,\mathrm{u} \)
\( \text{(B) Proton: } m = 1.00728\,\mathrm{u} \)
\( \text{(A) Electron: } m = 0.00055\,\mathrm{u} \)
The mass falls from (D) to (A), so the wavelength increases in the order D, C, B, A.