The kinetic energy of the electron, when the orbital radius of the electron in hydrogen atom is given as \( 0.53\,\mathrm{\mathring{A}} \), will be
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 13.6\,\mathrm{eV} \)
In the Bohr model of hydrogen, the radius \( 0.53\,\mathrm{\mathring{A}} \) is the Bohr radius \( r_1 \), so the electron is in the ground state (\( n = 1 \)).
The kinetic energy of the electron is equal in magnitude to its total energy:
\( \displaystyle K = \frac{e^{2}}{8\pi\varepsilon_0 r}, \qquad E = -\frac{e^{2}}{8\pi\varepsilon_0 r} \Rightarrow K = -E \)
\( E_1 = -13.6\,\mathrm{eV} \)
\( K = -E_1 = 13.6\,\mathrm{eV} \)
It can also be checked from the formula directly:
\( \displaystyle K = \frac{(9\times10^{9})(1.6\times10^{-19})^{2}}{2\times0.53\times10^{-10}}\,\mathrm{J} = 2.17\times10^{-18}\,\mathrm{J} \)
\( \displaystyle K = \frac{2.17\times10^{-18}}{1.6\times10^{-19}}\,\mathrm{eV} \approx 13.6\,\mathrm{eV} \)