| ⇦ | 
| ⇨ |
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: - Consider a two particle system with particle having masses m₁ and m₂. If the first
- Which wavelengths of sun is used finally as electric energy?
- A body starts from rest with an acceleration of 2 m/s². After 5 second,
- Two charges each equal to 2µ C are 0.5m apart. If both of them exist inside vacuum
- If an electron and a proton have the same de-Broglie wavelength, then the kinetic
Topics: Atoms and Nuclei
(136)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- Consider a two particle system with particle having masses m₁ and m₂. If the first
- Which wavelengths of sun is used finally as electric energy?
- A body starts from rest with an acceleration of 2 m/s². After 5 second,
- Two charges each equal to 2µ C are 0.5m apart. If both of them exist inside vacuum
- If an electron and a proton have the same de-Broglie wavelength, then the kinetic
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