| ⇦ | 
| ⇨ |
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.
Related Questions: - If a given sample of water contains 0.06 g of Ca²⁺ ion per kg, calculate
- Carborundum is
- Which of the following is a representation of Gay-Lussac’s law
- What is the value of carbonate hardness of water sample if 100 mL of it took
- A 0.66 kg ball is moving with a speed of 100 m/s. The associated wavelength
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 a given sample of water contains 0.06 g of Ca²⁺ ion per kg, calculate
- Carborundum is
- Which of the following is a representation of Gay-Lussac’s law
- What is the value of carbonate hardness of water sample if 100 mL of it took
- A 0.66 kg ball is moving with a speed of 100 m/s. The associated wavelength
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
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
Kinetic energy = negative of total energy or
-(-3.4)=3.4
I hope u like my explanation 😇