Chemistry Exam > Chemistry Questions > In hydrogen atom, energy of first excited sta...
Start Learning for Free
In hydrogen atom, energy of first excited state is - 3.4ev, then the K.E. of same orbital of hydrogen atom is ____ev
Correct answer is '3.4'. Can you explain this answer?
| FREE This question is part of | Download PDF Attempt this Test |
Most Upvoted Answer
In hydrogen atom, energy of first excited state is - 3.4ev, then the K...
Explanation:
The energy levels of an electron in a hydrogen atom are quantized, meaning they can only have certain specific values. These energy levels are given by the equation:
E = -13.6 eV / n^2
where E is the energy of the level and n is the principal quantum number. The energy of the first excited state is given as -3.4 eV, which corresponds to n = 2.
Calculating the Kinetic Energy:
The total energy of the electron in an orbital can be given as the sum of its kinetic energy (KE) and potential energy (PE). In the case of a hydrogen atom, the potential energy is negative and equal in magnitude to the total energy.
Total Energy (E) = KE + PE
Since the potential energy is equal to the total energy, we can rewrite the equation as:
E = KE + (-E)
Simplifying this equation, we get:
KE = -E - (-E)
= -E + E
= 0
Therefore, the kinetic energy of the electron in the first excited state of a hydrogen atom is 0 eV.
However, it's important to note that this result is specific to the first excited state. In general, the kinetic energy of an electron in a hydrogen atom can take on different values depending on the energy level it occupies.
The energy levels of an electron in a hydrogen atom are quantized, meaning they can only have certain specific values. These energy levels are given by the equation:
E = -13.6 eV / n^2
where E is the energy of the level and n is the principal quantum number. The energy of the first excited state is given as -3.4 eV, which corresponds to n = 2.
Calculating the Kinetic Energy:
The total energy of the electron in an orbital can be given as the sum of its kinetic energy (KE) and potential energy (PE). In the case of a hydrogen atom, the potential energy is negative and equal in magnitude to the total energy.
Total Energy (E) = KE + PE
Since the potential energy is equal to the total energy, we can rewrite the equation as:
E = KE + (-E)
Simplifying this equation, we get:
KE = -E - (-E)
= -E + E
= 0
Therefore, the kinetic energy of the electron in the first excited state of a hydrogen atom is 0 eV.
However, it's important to note that this result is specific to the first excited state. In general, the kinetic energy of an electron in a hydrogen atom can take on different values depending on the energy level it occupies.
Free Test
FREE
| Start Free Test |
Community Answer
In hydrogen atom, energy of first excited state is - 3.4ev, then the K...
Total energy = 1/2 potential energy
potential energy=2(-3.4)
kinetic energy=Total energy-potental energy=
-3.4-(-6.8)=3.4
potential energy=2(-3.4)
kinetic energy=Total energy-potental energy=
-3.4-(-6.8)=3.4
|
Explore Courses for Chemistry exam
|
|
Similar Chemistry Doubts
In hydrogen atom, energy of first excited state is - 3.4ev, then the K.E. of same orbital of hydrogen atom is ____evCorrect answer is '3.4'. Can you explain this answer?
Question Description
In hydrogen atom, energy of first excited state is - 3.4ev, then the K.E. of same orbital of hydrogen atom is ____evCorrect answer is '3.4'. Can you explain this answer? for Chemistry 2024 is part of Chemistry preparation. The Question and answers have been prepared according to the Chemistry exam syllabus. Information about In hydrogen atom, energy of first excited state is - 3.4ev, then the K.E. of same orbital of hydrogen atom is ____evCorrect answer is '3.4'. Can you explain this answer? covers all topics & solutions for Chemistry 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In hydrogen atom, energy of first excited state is - 3.4ev, then the K.E. of same orbital of hydrogen atom is ____evCorrect answer is '3.4'. Can you explain this answer?.
In hydrogen atom, energy of first excited state is - 3.4ev, then the K.E. of same orbital of hydrogen atom is ____evCorrect answer is '3.4'. Can you explain this answer? for Chemistry 2024 is part of Chemistry preparation. The Question and answers have been prepared according to the Chemistry exam syllabus. Information about In hydrogen atom, energy of first excited state is - 3.4ev, then the K.E. of same orbital of hydrogen atom is ____evCorrect answer is '3.4'. Can you explain this answer? covers all topics & solutions for Chemistry 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In hydrogen atom, energy of first excited state is - 3.4ev, then the K.E. of same orbital of hydrogen atom is ____evCorrect answer is '3.4'. Can you explain this answer?.
Solutions for In hydrogen atom, energy of first excited state is - 3.4ev, then the K.E. of same orbital of hydrogen atom is ____evCorrect answer is '3.4'. Can you explain this answer? in English & in Hindi are available as part of our courses for Chemistry.
Download more important topics, notes, lectures and mock test series for Chemistry Exam by signing up for free.
Here you can find the meaning of In hydrogen atom, energy of first excited state is - 3.4ev, then the K.E. of same orbital of hydrogen atom is ____evCorrect answer is '3.4'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
In hydrogen atom, energy of first excited state is - 3.4ev, then the K.E. of same orbital of hydrogen atom is ____evCorrect answer is '3.4'. Can you explain this answer?, a detailed solution for In hydrogen atom, energy of first excited state is - 3.4ev, then the K.E. of same orbital of hydrogen atom is ____evCorrect answer is '3.4'. Can you explain this answer? has been provided alongside types of In hydrogen atom, energy of first excited state is - 3.4ev, then the K.E. of same orbital of hydrogen atom is ____evCorrect answer is '3.4'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice In hydrogen atom, energy of first excited state is - 3.4ev, then the K.E. of same orbital of hydrogen atom is ____evCorrect answer is '3.4'. Can you explain this answer? tests, examples and also practice Chemistry tests.
|
|
Explore Courses for Chemistry exam
|
|
Suggested Free Tests
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
|
© EduRev
|
Education Revolution
|
Follow Us
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup