Arrange the following particles in the ascending order of their de-Broglie wavelengths.
A. An electron moving with velocity \(5\,\mathrm{ms^{-1}}\)
B. A proton moving with velocity \(10\,\mathrm{ms^{-1}}\)
C. A neutron moving with velocity \(20\,\mathrm{ms^{-1}}\)
D. An \(\alpha\)-particle moving with velocity \(15\,\mathrm{ms^{-1}}\)
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 \( \lambda = h/p = h/mv \), so the particle with the largest momentum has the shortest wavelength. Compare the momenta mv:
(A) An electron moving with velocity \(5\,\mathrm{ms^{-1}}\) — \( 9.1\times10^{-31} \times 5 \approx 4.6\times10^{-30}\,\mathrm{kg\,m\,s^{-1}} \)
(B) A proton moving with velocity \(10\,\mathrm{ms^{-1}}\) — \( 1.67\times10^{-27} \times 10 \approx 1.67\times10^{-26}\,\mathrm{kg\,m\,s^{-1}} \)
(C) A neutron moving with velocity \(20\,\mathrm{ms^{-1}}\) — \( 1.675\times10^{-27} \times 20 \approx 3.35\times10^{-26}\,\mathrm{kg\,m\,s^{-1}} \)
(D) An \(\alpha\)-particle moving with velocity \(15\,\mathrm{ms^{-1}}\) — \( 6.64\times10^{-27} \times 15 \approx 9.96\times10^{-26}\,\mathrm{kg\,m\,s^{-1}} \)
Momentum in decreasing order is D, C, B, A, so the wavelength in ascending order is D, C, B, A.
Shortcut: Take the proton (or neutron) mass as 1 unit: the momenta are about A: 0.003, B: 10, C: 20, D: 4 × 15 = 60, which gives the same order at once.