Question

The energy of the third orbit of \( {Li}^{2+} \) ion (in J) is:

• A. \( -2.18 \times 10^{-18} \, {J} \)
• B. \( -6.54 \times 10^{-18} \, {J} \)
• C. \( -7.3 \times 10^{-19} \, {J} \)
• D. \( +2.18 \times 10^{-18} \, {J} \)

Hint:

For hydrogen-like ions, use the formula \( E_n = - \dfrac{13.6 \, {eV}}{n^2} \times Z^2 \) to find the energy at different orbit levels.

Answer 1

The Correct Option is A

The energy of an electron in the \( n^{th} \) orbit of a hydrogen-like atom (or ion) is given by the formula: \[ E_n = - \dfrac{13.6 \, {eV}}{n^2} \] where \( n \) is the principal quantum number. For \( {Li}^{2+} \), the energy is given by: \[ E_n = - \dfrac{13.6 \, {eV}}{n^2} \times Z^2 \] where \( Z \) is the atomic number of \( {Li}^{2+} \), which is \( Z = 3 \). So, for the third orbit (\( n = 3 \)): \[ E_3 = - \dfrac{13.6 \times 3^2}{3^2} = - \dfrac{13.6}{9} \] This gives: \[ E_3 = - 1.51 \, {eV}. \] To convert this to joules, we multiply by \( 1.6 \times 10^{-19} \) (since \( 1 \, {eV} = 1.6 \times 10^{-19} \, {J} \)): \[ E_3 = - 1.51 \times 1.6 \times 10^{-19} \, {J} = - 2.18 \times 10^{-18} \, {J}. \] Thus, the energy of the third orbit of \( {Li}^{2+} \) ion is \( -2.18 \times 10^{-18} \, {J} \).