The potential energy of an electron in the first excited state of hydrogen atom is -6.8 eV. The kinetic energy of the electron in its ground state will be:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 13.6 eV
In a Bohr orbit the potential energy is \( U = -2K \), so the total energy is \( E = K + U = -K \). The kinetic energy is therefore the negative of the total energy of that orbit.
Check with the first excited state (\( n = 2 \)): \( K = 3.4\,\mathrm{eV} \) and \( U = -2K = -6.8\,\mathrm{eV} \), which matches the given value and gives \( E_2 = -3.4\,\mathrm{eV} = -13.6/2^{2}\,\mathrm{eV} \).
Ground state (\( n = 1 \))
\( E_1 = -\displaystyle\frac{13.6}{1^{2}}\,\mathrm{eV} = -13.6\,\mathrm{eV} \)
\( K_1 = -E_1 = +13.6\,\mathrm{eV} \)