The de Broglie wavelength of the electron in an orbit of a hydrogen atom is \(1.0\times10^{-9}\ \mathrm{m}\). What is the kinetic energy of the electron? (Take \(h=6.6\times10^{-34}\ \mathrm{J\,s}\), \(m=9.1\times10^{-31}\ \mathrm{kg}\) and \(1\ \mathrm{eV}=1.6\times10^{-19}\ \mathrm{J}\).)
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(1.5\ \mathrm{eV}\)
The de Broglie wavelength gives the momentum of the electron, \( p=h/\lambda \). Since \( p=mv \), the kinetic energy \( \frac{1}{2}mv^2 \) can be written as \( K=p^2/2m \).
Momentum of the electron:
\( \begin{aligned} p &= \frac{h}{\lambda} \\ &= \frac{6.6\times10^{-34}}{1.0\times10^{-9}} \\ &= 6.6\times10^{-25}\ \mathrm{kg\,m\,s^{-1}} \end{aligned} \)
Kinetic energy:
\( \begin{aligned} K &= \frac{p^2}{2m} \\ &= \frac{(6.6\times10^{-25})^2}{2\times9.1\times10^{-31}} \\ &= 2.4\times10^{-19}\ \mathrm{J} \end{aligned} \)
\( \begin{aligned} K &= \frac{2.4\times10^{-19}}{1.6\times10^{-19}} \\ &= 1.5\ \mathrm{eV} \end{aligned} \)
Therefore, the electron moving with a de Broglie wavelength of \( 1.0\times10^{-9}\ \mathrm{m} \) has a kinetic energy of about \( 1.5\ \mathrm{eV} \).
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.”