Question

Which of the following is correct match for hydrogen-like species for the total energy of e$^-$?

• A. 3rd orbit of Li$^{2+}$ ion = -21.6 × 10$^{-19}$ J/atom
• B. 2nd orbit of He$^{+}$ ion = -10.8 × 10$^{-19}$ J/atom
• C. 2nd orbit of Li$^{2+}$ ion = -9.6 × 10$^{-19}$ J/atom
• D. 2nd orbit of H-atom = -86.4 × 10$^{-19}$ J/atom

Hint:

The total energy of an electron in a hydrogen-like atom depends on both the atomic number \(Z\) and the principal quantum number \(n\). The energy is negative and decreases in magnitude as \(n\) increases.

Answer 1

The Correct Option is A

To determine which option correctly matches the total energy of an electron in hydrogen-like ions, we use the formula for the energy levels of such systems:

\(E_n = -\frac{13.6 \times Z^2}{n^2} \, \text{eV} \, \text{per electron}\)

Here, \(Z\) is the atomic number and \(n\) is the principal quantum number (orbit number). To convert from electronvolts to joules, we use the conversion factor: 1 eV = \(1.6 \times 10^{-19} \, \text{J}\).

1. For the 3rd orbit of \(\text{Li}^{2+}\) ion (Z = 3, n = 3):
   • Energy \(= -\frac{13.6 \times 3^2}{3^2} = -13.6 \, \text{eV}\)
   • Converting to joules: \(-13.6 \times 1.6 \times 10^{-19} = -21.76 \times 10^{-19} \, \text{J}\)
   • This value is approximately \(-21.6 \times 10^{-19} \, \text{J}\).
2. For the 2nd orbit of \(\text{He}^{+}\) ion (Z = 2, n = 2):
   • Energy \(= -\frac{13.6 \times 2^2}{2^2} = -13.6 \, \text{eV}\)
   • Converting to joules: \(-13.6 \times 1.6 \times 10^{-19} = -21.76 \times 10^{-19} \, \text{J}\)
   • This does not match the given value \(-10.8 \times 10^{-19} \, \text{J}\).
3. For the 2nd orbit of \(\text{Li}^{2+}\) ion (Z = 3, n = 2):
   • Energy \(= -\frac{13.6 \times 3^2}{2^2} = -30.6 \, \text{eV}\)
   • Converting to joules: \(-30.6 \times 1.6 \times 10^{-19} = -48.96 \times 10^{-19} \, \text{J}\)
   • This does not match the given value \(-9.6 \times 10^{-19} \, \text{J}\).
4. For the 2nd orbit of H-atom (Z = 1, n = 2):
   • Energy \(= -\frac{13.6 \times 1^2}{2^2} = -3.4 \, \text{eV}\)
   • Converting to joules: \(-3.4 \times 1.6 \times 10^{-19} = -5.44 \times 10^{-19} \, \text{J}\)
   • This does not match the given value \(-86.4 \times 10^{-19} \, \text{J}\).

Therefore, the correct answer is the 3rd orbit of \(\text{Li}^{2+}\) ion with energy \(-21.6 \times 10^{-19} \, \text{J/atom}\).