Arrange the de-Broglie wavelengths associated with the following in the decreasing order: (Given: \( h = 6.6 \times 10^{-34}\,\mathrm{Js} \))
(A) an electron (mass \( 9 \times 10^{-31}\,\mathrm{kg} \)) in the hydrogen atom moving with the speed of \( 3 \times 10^{6}\,\mathrm{m\,s^{-1}} \)
(B) a bullet (mass 0.04 kg) travelling at the speed of \( 10^{3}\,\mathrm{m\,s^{-1}} \)
(C) a ball (mass 0.06 kg) moving with the speed of \( 1.0\,\mathrm{m\,s^{-1}} \)
(D) a dust particle (mass \( 1.0 \times 10^{-9}\,\mathrm{kg} \)) drifting with the speed of \( 3.3\,\mathrm{m\,s^{-1}} \)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A), (D), (C), (B)
The de Broglie wavelength is \( \lambda = h/p = h/(mv) \). With \( h \) the same for all, the smaller the momentum, the longer the wavelength.
\( p_A = 9\times10^{-31} \times 3\times10^{6} = 2.7\times10^{-24}\,\mathrm{kg\,m\,s^{-1}} \)
\( p_B = 0.04 \times 10^{3} = 40\,\mathrm{kg\,m\,s^{-1}} \)
\( p_C = 0.06 \times 1.0 = 0.06\,\mathrm{kg\,m\,s^{-1}} \)
\( p_D = 1.0\times10^{-9} \times 3.3 = 3.3\times10^{-9}\,\mathrm{kg\,m\,s^{-1}} \)
\( \lambda_A = 6.6\times10^{-34}/2.7\times10^{-24} = 2.4\times10^{-10}\,\mathrm{m} \)
\( \lambda_D = 6.6\times10^{-34}/3.3\times10^{-9} = 2\times10^{-25}\,\mathrm{m} \)
\( \lambda_C = 6.6\times10^{-34}/0.06 = 1.1\times10^{-32}\,\mathrm{m} \)
\( \lambda_B = 6.6\times10^{-34}/40 = 1.65\times10^{-35}\,\mathrm{m} \)
\( (A)\ \text{electron}\ 2.4\times10^{-10} \gt (D)\ \text{dust particle}\ 2\times10^{-25} \gt (C)\ \text{ball}\ 1.1\times10^{-32} \gt (B)\ \text{bullet}\ 1.65\times10^{-35}\ (\mathrm{m}) \)
The decreasing order is A, D, C, B.
Shortcut: Since \( \lambda \propto 1/p \), only the momenta need comparing: \( p_A \lt p_D \lt p_C \lt p_B \) gives the order A, D, C, B directly.
Note: NTA picked this question from an NCERT (2026-27) exercise and changed the figures. Do not ignore NCERT while practising numericals.
NCERT (2026-27):
Exercises – …
