For hydrogen atom, energy of an electron in first excited state is – 3.4 eV, K.E. of the same electr — JEE Mains Chemistry Past Papers Chemistry Question

💡 Solution & Explanation

# Solution **Step 1: Identify the given information** - Energy of electron in first excited state (n=2): E₂ = –3.4 eV - Need to find: Kinetic Energy (K.E.) = x eV **Step 2: Apply the Virial Theorem** For hydrogen atom, the Virial Theorem states: - Total Energy (E) = –K.E. - Or: K.E. = –E **Step 3: Calculate Kinetic Energy** Since E₂ = –3.4 eV: K.E. = –E₂ = –(–3.4) = 3.4 eV **Step 4: Express in the required format** The answer should be in the form: x × 10⁻¹ eV Converting: 3.4 eV = 34 × 10⁻¹ eV **Step 5: Verify using alternative approach** Potential Energy in hydrogen: P.E. = 2E - P.E. = 2(–3.4) = –6.8 eV - K.E. = E – P.E. = –3.4 – (–6.8) = 3.4 eV ✓ Therefore, the answer is **34**