A bullet of mass 40 g is traveling at a speed of 1.0 km/s. Its de-Broglie wavelength would be
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 1.7 \times 10^{-35}\,\mathrm{m} \)
The de-Broglie wavelength of a particle of mass \( m \) moving with speed \( v \) is
\( \displaystyle \lambda = \frac{h}{p} = \frac{h}{m v} \)
\( m = 40\,\mathrm{g} = 0.040\,\mathrm{kg} \)
\( v = 1.0\,\mathrm{km\,s^{-1}} = 1000\,\mathrm{m\,s^{-1}} \)
\( h = 6.63\times10^{-34}\,\mathrm{J\,s} \)
\( \displaystyle \lambda = \frac{6.63\times10^{-34}}{0.040 \times 1000} = \frac{6.63\times10^{-34}}{40} = 1.66\times10^{-35}\,\mathrm{m} \approx 1.7\times10^{-35}\,\mathrm{m} \)
This is far smaller than any nuclear size, which is why the wave nature of a bullet is never noticed.
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):
Exercises – …
