| ⇦ |
| ⇨ |
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: - A long straight wire of radius a carries a steady current I. The current is uniformly
- The height at which the weight of a body becomes 1/16 th, its weight
- What is the value of incidence L for which the current is maximum in a series LCR
- A carnot engine working between 300 K and 600 K has a work output of 800J per cycle.
- In the diffraction pattern of a single slit
Topics: Atoms and Nuclei
(136)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A long straight wire of radius a carries a steady current I. The current is uniformly
- The height at which the weight of a body becomes 1/16 th, its weight
- What is the value of incidence L for which the current is maximum in a series LCR
- A carnot engine working between 300 K and 600 K has a work output of 800J per cycle.
- In the diffraction pattern of a single slit
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