An electron moving with velocity of \( 5.4\times10^{6}\,\mathrm{m/sec} \). What is the de-Broglie wavelength associated with them? \( (h=6.63\times10^{-34}\,\mathrm{J\,s}) \)
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 1.35 Å
The de Broglie wavelength of a particle of mass m moving with speed v is \( \lambda = h/(mv) \). For an electron \( m = 9.1\times10^{-31}\,\mathrm{kg} \).
\( \displaystyle \lambda = \frac{6.63\times10^{-34}}{9.1\times10^{-31}\times5.4\times10^{6}} = \frac{6.63\times10^{-34}}{4.914\times10^{-24}}\,\mathrm{m} \)
\( \lambda \approx 1.35\times10^{-10}\,\mathrm{m} = 1.35\,\mathrm{\mathring{A}} \)
Note: This question is from NCERT (2026-27). NTA sets numerical questions from NCERT too, so do not ignore NCERT while practising numericals.
NCERT (2026-27):
Wave Nature of Matter – …
