| ⇦ | 
| ⇨ |
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:
- Energy stored in the coil of a self inductance 40 mH carrying a steady current of 2A
- Two balls are dropped from heights h and 2h respectively. The ratio of times
- A body of mass (4m) is lying in x-y plane at rest. It suddenly explodes into three
- Which of the following phenomenon exhibited the particle nature of light?
- The drift velocity of free electrons in a conductor is v, when a current
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