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The total energy of an electron in the first excited state of hydrogen atom is about -3.4 eV. Its kinetic energy in this state is
Options
(a) 3.4 eV
(b) 6.8 eV
(c) -3.4 eV
(d) -6.8 eV
Correct Answer:
3.4 eV
Explanation:
KE. = |(1/2) P.E.|
But P.E. is negavite
.·. Total energy = |(1/2) P.E.| – P.E. = – P.E. / 2 = – 3.4 eV
.·. K.E. = + 3.4 eV
Related Questions: - Given the value of Rydberg constant is 10⁷ m⁻¹, the wave number of the last line
- Excitation energy of a hydrogen like ion, in its first excitation state, is 40.8 eV.
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Topics: Atoms and Nuclei
(136)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- Given the value of Rydberg constant is 10⁷ m⁻¹, the wave number of the last line
- Excitation energy of a hydrogen like ion, in its first excitation state, is 40.8 eV.
- A sample of HCl gas is placed in an electric field of 5 x 10⁴ N/C
- In cyclotron, for a given magnet, radius of the semicircle traced by positive ion
- Power dissipated in an LCR series circuit connected to an a.c source of emf ? is
Topics: Atoms and Nuclei (136)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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