In the first excited state of hydrogen atom, the energy of its electron is -3.4 ~eV$-3.4 \mathrm{~eV}$. The radial distance of the electron from the hydrogen nucleus in this case is approximately: (Take 1eV = 1.6 × 10⁻¹⁹ ~J$1\mathrm{eV} = 1.6 \times 10^{-19} \mathrm{~J}$, e = 1.6 × 10⁻¹⁹ ~C$e = 1.6 \times 10^{-19} \mathrm{~C}$ and (1)/(4πε₀) = 9 × 10⁹ ~N m²/C²$\frac{1}{4\pi\varepsilon_0} = 9 \times 10^9 \mathrm{~N\ m^2/C^2}$)

Related Formula

Core Logic

Given Total Energy E = -3.4 ~eV$E = -3.4 \mathrm{~eV}$. The Kinetic Energy is exactly the positive magnitude of the total energy: K.E. = +3.4 ~eV$K.E. = +3.4 \mathrm{~eV}$.

The electrostatic force provides the necessary centripetal force:

Since K.E. = (1)/(2) m v²$K.E. = \frac{1}{2} m v^2$, substituting yields:

Step 1: Compute Radial Distance

Equating the two kinetic energy expressions:

Pattern Recognition

Bohr radius shortcut: rₙ = 0.529 × n² ~AA$r_n = 0.529 \times n^2 \mathrm{~\AA}$. First excited state is n=2$n=2$, meaning r₂ = 0.529 × 4 = 2.116 ~AA = 2.1 × 10⁻¹⁰ ~m$r_2 = 0.529 \times 4 = 2.116 \mathrm{~\AA} = 2.1 \times 10^{-10} \mathrm{~m}$.

Chapter Mix

Class 12 Physics: Atoms