The ionisation energy of the hydrogen atom is 13.6 eV. Following Bohr’s theory, the energy corresponding to a transition between the 3rd and the 4th orbit is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 0.66 eV
In the Bohr model the energy of the electron in the \( n \)th orbit is \( E_n=-13.6\,\mathrm{eV}/n^2 \). The ionisation energy 13.6 eV is the energy needed to take the electron from \( E_1=-13.6\,\mathrm{eV} \) to zero. The energy of a transition is the difference of the two orbit energies.
\( \begin{aligned} E_3 &= -\frac{13.6}{3^2}\,\mathrm{eV} \\ &= -1.51\,\mathrm{eV} \end{aligned} \)
\( \begin{aligned} E_4 &= -\frac{13.6}{4^2}\,\mathrm{eV} \\ &= -0.85\,\mathrm{eV} \end{aligned} \)
\( \begin{aligned} \Delta E &= E_4-E_3 \\ &= -0.85\,\mathrm{eV}-(-1.51\,\mathrm{eV}) \\ &= 0.66\,\mathrm{eV} \end{aligned} \)