A hydrogen atom in its first excited state absorbs a photon of energy x \( \times \) 10\( - \)2 eV and excited to a higher energy state where the potential energy of electron is \( - \)1.08 eV. The value of x is ______________.
A hydrogen atom in its first excited state absorbs a photon of energy x \( \times \) 10\( - \)2 eV and excited to a higher energy state where the potential energy of electron is \( - \)1.08 eV. The value of x is ______________.
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Correct Answer: 286
Solution and Explanation
Step 2: En = -13.6/n2 = -0.54 gives n2 \(\approx\) 25, n = 5 (E5 = -13.6/25 = -0.544 eV).
Step 3: First excited state is n = 2: E2 = -3.4 eV.
Step 4: Photon energy = E5 - E2 = -0.544 + 3.4 = 2.856 eV = 285.6 × 10-2 eV \(\approx\) 286 × 10-2 eV.
Step 5: So x \(\approx\) 286.
Answer: 286
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