The electron in a hydrogen atom revolves in a circular orbit with a speed of \(1.1\times10^{6}\ \mathrm{m\,s^{-1}}\). The total energy of the electron in this orbit is about (Take mass of electron \(m=9.1\times10^{-31}\ \mathrm{kg}\) and \(1\ \mathrm{eV}=1.6\times10^{-19}\ \mathrm{J}\).)
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(-3.4\ \mathrm{eV}\)
For the electron in a hydrogen orbit, the kinetic energy is \( K=e^{2}/8\pi\varepsilon_0r \) and the total energy is \( E=-e^{2}/8\pi\varepsilon_0r \). So the total energy is the negative of the kinetic energy, \( E=-K \).
\( \begin{aligned} K &= \frac{1}{2}mv^{2} \\ &= \frac{1}{2}\times9.1\times10^{-31}\times(1.1\times10^{6})^{2} \\ &= 5.5\times10^{-19}\ \mathrm{J} \end{aligned} \)
\( \begin{aligned} K &= \frac{5.5\times10^{-19}}{1.6\times10^{-19}}\ \mathrm{eV} \\ &= 3.4\ \mathrm{eV} \end{aligned} \)
\( \begin{aligned} E &= -K \\ &= -3.4\ \mathrm{eV} \end{aligned} \)
The negative sign shows that the electron is bound to the nucleus. Hence, the total energy of the electron is about \( -3.4\ \mathrm{eV} \).
NCERT (2026-27):
Electron orbits – “The kinetic energy (K ) and electrostatic potential energy (U) of the electron in hydrogen atom are”
\( \displaystyle K=\frac{1}{2}mv^{2}=\frac{e^{2}}{8\pi\varepsilon_0r} \)\( \quad \text{and} \quad \displaystyle U=-\frac{e^{2}}{4\pi\varepsilon_0r} \)
Electron orbits – “… Thus the total energy E of the electron in a hydrogen atom is”
\( \displaystyle E=K+U=\frac{e^{2}}{8\pi\varepsilon_0r}-\frac{e^{2}}{4\pi\varepsilon_0r} \)
\( \displaystyle E=-\frac{e^{2}}{8\pi\varepsilon_0r} \)\( \qquad (12.4) \)