Related Formula
Eₙ = -13.6 (Z²)/(n²)~eV$$E_n = -13.6 \frac{Z^2}{n^2}\mathrm{~eV}$$
where Z$Z$ is the atomic number and n$n$ is the principal quantum number.
Core Logic
Let's check each statement by calculating energy values:
- H atom ground state (Z = 1, n = 1$Z = 1, n = 1$):
EH = -13.6 (1²)/(1²) = -13.6~eV$$E_H = -13.6 \frac{1^2}{1^2} = -13.6\mathrm{~eV}$$
- Statement (A): He^+$\mathrm{He}^+$ ion in its first excited state (Z = 2, n = 2$Z = 2, n = 2$):
EHe^+, n=2 = -13.6 (2²)/(2²) = -13.6~eV$$E_{\mathrm{He}^+, n=2} = -13.6 \frac{2^2}{2^2} = -13.6\mathrm{~eV}$$
This is equal to EH$E_H$. Thus, Statement (A) is correct.
- Statement (B): Li⁺⁺$\mathrm{Li}^{++}$ ion in its second excited state (Z = 3, n = 3$Z = 3, n = 3$):
ELi⁺⁺, n=3 = -13.6 (3²)/(3²) = -13.6~eV$$E_{\mathrm{Li}^{++}, n=3} = -13.6 \frac{3^2}{3^2} = -13.6\mathrm{~eV}$$
This is equal to EH$E_H$. Thus, Statement (B) is correct.
- Statement (C): He^+$\mathrm{He}^+$ ion in its ground state (Z = 2, n = 1$Z = 2, n = 1$):
EHe^+, n=1 = -13.6 (2²)/(1²) = -54.4~eV ≠ EH$$E_{\mathrm{He}^+, n=1} = -13.6 \frac{2^2}{1^2} = -54.4\mathrm{~eV} \neq E_H$$
Thus, Statement (C) is incorrect.
- Statement (D): He^+$\mathrm{He}^+$ in first excited state (Z=2, n=2$Z=2, n=2$) versus Li⁺⁺$\mathrm{Li}^{++}$ in ground state (Z=3, n=1$Z=3, n=1$):
ELi⁺⁺, n=1 = -13.6 (3²)/(1²) = -122.4~eV ≠ -13.6~eV$$E_{\mathrm{Li}^{++}, n=1} = -13.6 \frac{3^2}{1^2} = -122.4\mathrm{~eV} \neq -13.6\mathrm{~eV}$$
Thus, Statement (D) is incorrect.
Step 1: Final Conclusion
Statements (A) and (B) are the only correct statements.
Pattern Recognition
Bohr energy is proportional to (Z²)/(n²)$\frac{Z^2}{n^2}$. If the ratio of (Z)/(n)$\frac{Z}{n}$ is the same for two systems, their quantum state energy levels will be identical.
Chapter Mix
Class 12 Physics: Atoms
