Assume that the following four particles have same de-Broglie wavelength. Arrange them in increasing order of their Kinetic Energies (K.E.).
(A) \( E_{1} \) : K.E. of electron
(B) \( E_{2} \) : K.E. of deuteron
(C) \( E_{3} \) : K.E. of proton
(D) \( E_{4} \) : K.E. of neutron
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (B), (D), (C), (A)
By the de Broglie relation \( \lambda = h/p \), equal wavelengths mean equal momenta. For a given momentum, the kinetic energy is inversely proportional to the mass:
\( \displaystyle K = \frac{p^{2}}{2m} \ \Rightarrow\ K \propto \frac{1}{m} \)
Masses:
\( \text{deuteron} \approx 2.014\,\mathrm{u} \gt \text{neutron} = 1.0087\,\mathrm{u} \gt \text{proton} = 1.0073\,\mathrm{u} \gt \text{electron} \approx 0.00055\,\mathrm{u} \)
The heaviest particle has the least kinetic energy, so in increasing order of K.E.:
\( E_{2}\ \text{(deuteron)} \lt E_{4}\ \text{(neutron)} \lt E_{3}\ \text{(proton)} \lt E_{1}\ \text{(electron)} \)