The ground state energy of a hydrogen atom is -13.6 eV. The energy of second excited state of He+ ion in eV is:

🧠 Energy Level Decoding In Bohr's model, energy is quantized and depends on the nuclear pull ($Z$) and the distance ($n$). The term "2nd excited state" is a common trap—it refers to the third energy level ($n=3$). We need to find the energy for $\ce{He^+}$ at this level.

🗺️ The Molar Multiplier

1. The Formula: $E_n = -13.6 \times \frac{Z^2}{n^2} \text{ eV}$.
2. The Variables:
   • For $\ce{He^+}$, $Z = 2$.
   • For the 2nd excited state, $n = 3$.
3. Calculation: $E_3 = -13.6 \times \frac{2^2}{3^2} = -13.6 \times \frac{4}{9}$ $E_3 \approx -1.51 \times 4 = -6.04 \text{ eV}$

⚡ The $1.51$ Shortcut Remember the Hydrogen energy levels: $13.6, 3.4, 1.51$. Since $E(\ce{He^+}) = Z^2 \times E(\text{H})$, just multiply the Hydrogen $n=3$ energy by $4$. $1.51 \times 4 = 6.04 \text{ eV}$. This bypasses the $13.6/9$ division entirely!

⚠️ Common Traps Mixing up the "excited state" number.

• Ground state $= n=1$
• 1st excited state $= n=2$
• 2nd excited state $= n=3$ Missing this counts for $90$ of errors in this question type.

$\boxed{\text{Answer: (b)}}$