The kinetic energy of an electron in the H-atom in the second excited state is \( k \) units. Its total energy and potential energy, respectively, in the ground state will be
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( -9k, -18k \)
In the Bohr model of hydrogen the energies of an orbit are related by
\( \displaystyle K = -E, \qquad U = 2E = -2K, \qquad E_n = -\frac{13.6}{n^{2}}\,\mathrm{eV} \)
So the kinetic energy is proportional to \( 1/n^{2} \). The second excited state is \( n = 3 \) and the ground state is \( n = 1 \).
\( \displaystyle \frac{K_1}{K_3} = \frac{3^{2}}{1^{2}} = 9 \;\Rightarrow\; K_1 = 9k \)
\( E_1 = -K_1 = -9k \)
\( U_1 = -2K_1 = -18k \)