The total energy of the electron in an orbit of a hydrogen atom is \(-1.51\ \mathrm{eV}\). The radius of this orbit is about (Take \(1/4\pi\varepsilon_0=9\times10^{9}\ \mathrm{N\,m^{2}\,C^{-2}}\) and \(e=1.6\times10^{-19}\ \mathrm{C}\).)
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(4.8\times10^{-10}\ \mathrm{m}\)
In a hydrogen atom, an electron moving in an orbit of radius \( r \) has energy \( E=-e^{2}/8\pi\varepsilon_0r \). The negative sign shows that the electron is bound to the nucleus. So the radius is \( r=-e^{2}/8\pi\varepsilon_0E \).
First, change the energy into joules.
\( \begin{aligned} E &= -1.51\times1.6\times10^{-19}\ \mathrm{J} \\ &= -2.42\times10^{-19}\ \mathrm{J} \end{aligned} \)
\( \begin{aligned} r &= -\frac{1}{4\pi\varepsilon_0}\cdot\frac{e^{2}}{2E} \\ &= -\frac{9\times10^{9}\times(1.6\times10^{-19})^{2}}{2\times(-2.42\times10^{-19})} \\ &= 4.8\times10^{-10}\ \mathrm{m} \end{aligned} \)
Therefore, the radius of the orbit is about \( 4.8\times10^{-10}\ \mathrm{m} \).
NCERT (2026-27):
Electron orbits – “… Thus the total energy E of the electron in a hydrogen atom is”
\( \displaystyle E=K+U=\frac{e^{2}}{8\pi\varepsilon_0r}-\frac{e^{2}}{4\pi\varepsilon_0r} \)
\( \displaystyle E=-\frac{e^{2}}{8\pi\varepsilon_0r} \)\( \qquad (12.4) \)