The ionization energy of a hydrogen-like atom is 14.4 eV. The amount of energy released when electro — Atomic Structure Chemistry Question
Question
The ionization energy of a hydrogen-like atom is 14.4 eV. The amount of energy released when electron jumps from the fourth orbit to the first orbit in this atom, is
Answer: A
💡 Solution & Explanation
Ionization energy is 14.4 eV, meaning the ground state energy $E_1 = -14.4$ eV. Energy released $\Delta E = E_4 - E_1 = \left(\frac{-14.4}{4^2}\right) - (-14.4) = -0.9 + 14.4 = 13.5$ eV. Therefore, correct answer is A.
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