| ⇦ | 
| ⇨ |
The excitation potential of hydrogen atom in the first excited state is
Options
(a) 13.6 V
(b) (-13.6 V)
(c) 10.2 V
(d) (-10 V)
Correct Answer:
10.2 V
Explanation:
Energy of electron in ground state of hydrogen atom,
E₁ = -1.36 eV [E(n) = (-13.6 / n²) eV]
Energy of electron in first excited state of hydrogen atom,
E₂ = -13.6 / 2² = – 3.4 eV
Hence, the excitation energy of hydrogen atom in first excited state,
E = E₂ – E₁ = – 3.4 – (-13.6) = 10.2 eV
Therefore, excitation potential is 10.2 V.
Related Questions: - If μᵥ=1.5230 and μʀ=15.145, then dispersive power of crown glass is
- In potentiometer experiment, a cell of emf 1.25V gives balancing length of 30 cm
- The ionization energy of Li⁺⁺ is equal to
- A super conductor exhibits perfect
- Under constant pressure, graph between P and 1/V is a
Topics: Atoms and Nuclei
(136)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- If μᵥ=1.5230 and μʀ=15.145, then dispersive power of crown glass is
- In potentiometer experiment, a cell of emf 1.25V gives balancing length of 30 cm
- The ionization energy of Li⁺⁺ is equal to
- A super conductor exhibits perfect
- Under constant pressure, graph between P and 1/V is a
Topics: Atoms and Nuclei (136)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
18000+ students are using NEETLab to improve their score. What about you?
Solve Previous Year MCQs, Mock Tests, Topicwise Practice Tests, Identify Weak Topics, Formula Flash cards and much more is available in NEETLab Android App to improve your NEET score.
Share this page with your friends
Leave a Reply