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

Step 1: Total energy of electron in hydrogen-like species.
The total energy \( E \) of an electron in a hydrogen-like atom is given by the equation: \[ E = -21.6 \times 10^{-19} \times \frac{Z^2}{n^2} \, \text{J/atom} \] Where \( Z \) is the atomic number and \( n \) is the principal quantum number of the orbit.
Step 2: Analyzing the options.
For the 3rd orbit of Li$^{2+}$, \( Z = 3 \) and \( n = 3 \), thus: \[ E = -21.6 \times 10^{-19} \times \frac{3^2}{3^2} = -21.6 \times 10^{-19} \, \text{J/atom} \] This matches option (1).
Step 3: Conclusion.
The correct answer is (1), as the calculation matches the given total energy.