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In hydrogen atom, energy of first excited state is –3.4 eV. Find out KE of the same orbit of Hydrogen atom
Options
(a) +3.4 eV
(b) +6.8 eV
(c) –13.6 eV
(d) +13.6 eV
Correct Answer:
+3.4 eV
Explanation:
No explanation available. Be the first to write the explanation for this question by commenting below.
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Question Type: Memory
(964)
Difficulty Level: Easy
(1008)
Topics: Structure of Atom
(90)
Subject: Chemistry
(2512)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- If heat of formation of C + O₂ → 2CO is ΔH = -110.5 kcal,then heat of combustion
- In methane four C-H bonds are directed towards the corners of
- Correct relation between dissociation constant of a dibasic acid is
- The maximum number of electrons in a subshell is given by the expression
- During the decomposition of H₂O₂ to give oxygen, 48 g O₂ is formed per minute
Question Type: Memory (964)
Difficulty Level: Easy (1008)
Topics: Structure of Atom (90)
Subject: Chemistry (2512)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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Kinetic energy = negative of total energy or
-(-3.4)=3.4
I hope u like my explanation 😇