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
Question
The energy of an electron in first Bohr orbit of H–atom is –13.6 eV. The magnitude of energy value of electron in the first excited state of Be3+ is ___________ eV. (nearest integer value)
💡 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**