In a hydrogen atom, the electron moves in an orbit of radius \(2.6\times10^{-9}\ \mathrm{m}\). Seven de Broglie wavelengths of the electron fit into the circumference of this orbit. The speed of the electron is about (Take \(h=6.6\times10^{-34}\ \mathrm{J\,s}\) and \(m=9.1\times10^{-31}\ \mathrm{kg}\).)
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(3.1\times10^{5}\ \mathrm{m\,s^{-1}}\)
In a Bohr orbit the electron wave forms a standing wave. So the circumference holds a whole number of de Broglie wavelengths, here 7. The speed then comes from \( \lambda=h/mv \).
De Broglie wavelength:
\( \begin{aligned} \lambda &= \frac{2\pi r}{7} \\ &= \frac{2\times3.14\times2.6\times10^{-9}}{7} \\ &= 2.33\times10^{-9}\ \mathrm{m} \end{aligned} \)
Speed of the electron:
\( \begin{aligned} v &= \frac{h}{m\lambda} \\ &= \frac{6.6\times10^{-34}}{9.1\times10^{-31}\times2.33\times10^{-9}} \\ &= 3.1\times10^{5}\ \mathrm{m\,s^{-1}} \end{aligned} \)
From the above, \( v=3.1\times10^{5}\ \mathrm{m\,s^{-1}} \). Therefore, the electron moves at about \( 3.1\times10^{5}\ \mathrm{m\,s^{-1}} \) in this orbit.
NCERT (2026-27):
Summary – “8. de Broglie’s hypothesis that electrons have a wavelength λ = h/mv gave an explanation for Bohr’s quantised orbits by bringing in the wave-particle duality. The orbits correspond to circular standing waves in which the circumference of the orbit equals a whole number of wavelengths.”