The kinetic energy of an electron in the 2nd excited state in hydrogen atom, if its ground state energy is -13.6 eV, will be
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 1.51 eV
In a Bohr orbit the kinetic energy is the negative of the total energy, \( K = -E_n \), where \( E_n = -13.6/n^{2}\,\mathrm{eV} \). The ground state is \( n = 1 \), the first excited state \( n = 2 \), and the 2nd excited state is \( n = 3 \).
\( \displaystyle E_{3} = -\frac{13.6}{3^{2}} = -1.51\,\mathrm{eV} \)
\( \displaystyle K = -E_{3} = 1.51\,\mathrm{eV} \)