The de Broglie wavelength of an electron is 0.33 nm. The speed of the electron is (Take \(h = 6.6\times10^{-34}\ \mathrm{J\,s}\) and mass of the electron \(= 9.0\times10^{-31}\ \mathrm{kg}\).)
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(2.2\times10^{6}\ \mathrm{m\,s^{-1}}\)
For a particle of mass \( m \) and speed \( v \), the de Broglie relation is \( \lambda=h/mv \). Rearranged, it gives the speed as \( v=h/m\lambda \).
\( \lambda=0.33\ \mathrm{nm}=3.3\times10^{-10}\ \mathrm{m} \)
\( \begin{aligned} v &= \frac{h}{m\lambda} \\ &= \frac{6.6\times10^{-34}}{9.0\times10^{-31}\times3.3\times10^{-10}} \\ &= 2.2\times10^{6}\ \mathrm{m\,s^{-1}} \end{aligned} \)
Hence, the electron moves at \( 2.2\times10^{6}\ \mathrm{m\,s^{-1}} \).
NCERT (2026-27):
Wave Nature of Matter – “… De Broglie proposed that the wave length λ associated with a particle of momentum p is given as”
\( \displaystyle \lambda=\frac{h}{p}=\frac{h}{mv} \)\( \qquad (11.5) \)
“where m is the mass of the particle and v its speed.”