| ⇦ | 
| ⇨ |
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: - The temperature at which the vapour pressure of a liquid becomes equal to the external
- A parallel beam of light of wavelength λ is incident normally on a narrow slit
- A p-n photodiode is made of a material with a band gap of 2.0 eV
- A particle of mass m executes simple harmonic motion with amplitude
- If angular momentum of a body is increased by 200% its kinetic energy will increase
Topics: Atoms and Nuclei
(136)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- The temperature at which the vapour pressure of a liquid becomes equal to the external
- A parallel beam of light of wavelength λ is incident normally on a narrow slit
- A p-n photodiode is made of a material with a band gap of 2.0 eV
- A particle of mass m executes simple harmonic motion with amplitude
- If angular momentum of a body is increased by 200% its kinetic energy will increase
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