The energy of an electron in first Bohr orbit of H–atom is –13.6 eV. The magnitude of energy value o — JEE Mains Chemistry Past Papers Chemistry Question

💡 Solution & Explanation

# Solution **Step 1: Identify the relevant formula** For hydrogen-like ions (single electron), the energy of an electron in the nth orbit is: $$E_n = -13.6 \times \frac{Z^2}{n^2} \text{ eV}$$ where Z is the atomic number and n is the principal quantum number. **Step 2: Identify the given information** - For H-atom (Z = 1, n = 1): E₁ = –13.6 eV ✓ (confirms the formula) - For Be³⁺: Z = 4 (beryllium has 4 protons) - First excited state means n = 2 (the electron is promoted from n = 1) **Step 3: Calculate energy for Be³⁺ in first excited state** $$E_2 = -13.6 \times \frac{4^2}{2^2}$$ $$E_2 = -13.6 \times \frac{16}{4}$$ $$E_2 = -13.6 \times 4 = -54.4 \text{ eV}$$ **Step 4: Find the magnitude** Magnitude of energy = |–54.4| = 54.4 eV **Step 5: Round to nearest integer** 54.4 eV ≈ 54 eV Therefore, the answer is **54 eV**