Chemistry Exam > Chemistry Questions > Assume that for a real system, a real wave fu...
Start Learning for Free
Assume that for a real system, a real wave function is a linear combination of two orthonormal basis functions where energy integrals are H11 = –15, H22 = –4, H12 = H21 = –1. What would be the ground state and excited wave functions?
- a)G.S.(ᵹ) = 0.996 Փ1 + 0.0896Փ2 and E.S.(ᵹ) = –0.996 Փ1 + 0.0896Փ2
- b)G.S.(ᵹ) = 0.996 Փ1 + 0.0896Փ2 and E.S.(ᵹ) = –0.0896 Փ1 + 0.996Փ2.
- c)G.S.(ᵹ) = –15.09 Փ1 – 3.91Փ2 and E.S.(ᵹ) = –3.91 Փ1 – 15.09Փ2
- d)G.S.(ᵹ) = 15.09 Փ1 + 3.91Փ2 and E.S.(ᵹ) = –3.91 Փ1 – 15.09Փ2
Correct answer is option 'B'. Can you explain this answer?
| FREE This question is part of | Download PDF Attempt this Test |
Most Upvoted Answer
Assume that for a real system, a real wave function is a linear combin...
E11, H22 = E22, and H12 = H21 = 0. In other words, the Hamiltonian matrix for this system is given by:
H = | E11 0 |
| 0 E22 |
Here, E11 and E22 represent the energies associated with the two basis functions, and the off-diagonal elements of the Hamiltonian matrix are zero.
To find the energy eigenvalues and eigenfunctions for this system, we need to solve the Schrödinger equation:
Hψ = Eψ
Plugging in the Hamiltonian matrix, we get:
| E11 0 | | ψ1 | = E | ψ1 |
| 0 E22 | | ψ2 | | ψ2 |
This equation can be written as two separate equations:
E11ψ1 = Eψ1
E22ψ2 = Eψ2
Simplifying, we see that the eigenvalues E are simply equal to the energies E11 and E22 associated with the basis functions. The corresponding eigenfunctions ψ1 and ψ2 are the coefficients of the basis functions in the linear combination.
In other words, the energy eigenvalues for this system are E11 and E22, and the corresponding eigenfunctions are the two orthonormal basis functions.
H = | E11 0 |
| 0 E22 |
Here, E11 and E22 represent the energies associated with the two basis functions, and the off-diagonal elements of the Hamiltonian matrix are zero.
To find the energy eigenvalues and eigenfunctions for this system, we need to solve the Schrödinger equation:
Hψ = Eψ
Plugging in the Hamiltonian matrix, we get:
| E11 0 | | ψ1 | = E | ψ1 |
| 0 E22 | | ψ2 | | ψ2 |
This equation can be written as two separate equations:
E11ψ1 = Eψ1
E22ψ2 = Eψ2
Simplifying, we see that the eigenvalues E are simply equal to the energies E11 and E22 associated with the basis functions. The corresponding eigenfunctions ψ1 and ψ2 are the coefficients of the basis functions in the linear combination.
In other words, the energy eigenvalues for this system are E11 and E22, and the corresponding eigenfunctions are the two orthonormal basis functions.
|
Explore Courses for Chemistry exam
|
|
Assume that for a real system, a real wave function is a linear combination of two orthonormal basis functions where energy integrals are H11 = –15, H22 = –4, H12 = H21 = –1. What would be the ground state and excited wave functions?a)G.S.() = 0.996 1 + 0.08962 and E.S.() = –0.996 1 + 0.08962b)G.S.() = 0.996 1 + 0.08962 and E.S.() = –0.0896 1 + 0.9962.c)G.S.() = –15.09 1 – 3.912 and E.S.() = –3.91 1 – 15.092d)G.S.() = 15.09 1 + 3.912 and E.S.() = –3.91 1 – 15.092Correct answer is option 'B'. Can you explain this answer?
Question Description
Assume that for a real system, a real wave function is a linear combination of two orthonormal basis functions where energy integrals are H11 = –15, H22 = –4, H12 = H21 = –1. What would be the ground state and excited wave functions?a)G.S.() = 0.996 1 + 0.08962 and E.S.() = –0.996 1 + 0.08962b)G.S.() = 0.996 1 + 0.08962 and E.S.() = –0.0896 1 + 0.9962.c)G.S.() = –15.09 1 – 3.912 and E.S.() = –3.91 1 – 15.092d)G.S.() = 15.09 1 + 3.912 and E.S.() = –3.91 1 – 15.092Correct answer is option 'B'. 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 Assume that for a real system, a real wave function is a linear combination of two orthonormal basis functions where energy integrals are H11 = –15, H22 = –4, H12 = H21 = –1. What would be the ground state and excited wave functions?a)G.S.() = 0.996 1 + 0.08962 and E.S.() = –0.996 1 + 0.08962b)G.S.() = 0.996 1 + 0.08962 and E.S.() = –0.0896 1 + 0.9962.c)G.S.() = –15.09 1 – 3.912 and E.S.() = –3.91 1 – 15.092d)G.S.() = 15.09 1 + 3.912 and E.S.() = –3.91 1 – 15.092Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for Chemistry 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Assume that for a real system, a real wave function is a linear combination of two orthonormal basis functions where energy integrals are H11 = –15, H22 = –4, H12 = H21 = –1. What would be the ground state and excited wave functions?a)G.S.() = 0.996 1 + 0.08962 and E.S.() = –0.996 1 + 0.08962b)G.S.() = 0.996 1 + 0.08962 and E.S.() = –0.0896 1 + 0.9962.c)G.S.() = –15.09 1 – 3.912 and E.S.() = –3.91 1 – 15.092d)G.S.() = 15.09 1 + 3.912 and E.S.() = –3.91 1 – 15.092Correct answer is option 'B'. Can you explain this answer?.
Assume that for a real system, a real wave function is a linear combination of two orthonormal basis functions where energy integrals are H11 = –15, H22 = –4, H12 = H21 = –1. What would be the ground state and excited wave functions?a)G.S.() = 0.996 1 + 0.08962 and E.S.() = –0.996 1 + 0.08962b)G.S.() = 0.996 1 + 0.08962 and E.S.() = –0.0896 1 + 0.9962.c)G.S.() = –15.09 1 – 3.912 and E.S.() = –3.91 1 – 15.092d)G.S.() = 15.09 1 + 3.912 and E.S.() = –3.91 1 – 15.092Correct answer is option 'B'. 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 Assume that for a real system, a real wave function is a linear combination of two orthonormal basis functions where energy integrals are H11 = –15, H22 = –4, H12 = H21 = –1. What would be the ground state and excited wave functions?a)G.S.() = 0.996 1 + 0.08962 and E.S.() = –0.996 1 + 0.08962b)G.S.() = 0.996 1 + 0.08962 and E.S.() = –0.0896 1 + 0.9962.c)G.S.() = –15.09 1 – 3.912 and E.S.() = –3.91 1 – 15.092d)G.S.() = 15.09 1 + 3.912 and E.S.() = –3.91 1 – 15.092Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for Chemistry 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Assume that for a real system, a real wave function is a linear combination of two orthonormal basis functions where energy integrals are H11 = –15, H22 = –4, H12 = H21 = –1. What would be the ground state and excited wave functions?a)G.S.() = 0.996 1 + 0.08962 and E.S.() = –0.996 1 + 0.08962b)G.S.() = 0.996 1 + 0.08962 and E.S.() = –0.0896 1 + 0.9962.c)G.S.() = –15.09 1 – 3.912 and E.S.() = –3.91 1 – 15.092d)G.S.() = 15.09 1 + 3.912 and E.S.() = –3.91 1 – 15.092Correct answer is option 'B'. Can you explain this answer?.
Solutions for Assume that for a real system, a real wave function is a linear combination of two orthonormal basis functions where energy integrals are H11 = –15, H22 = –4, H12 = H21 = –1. What would be the ground state and excited wave functions?a)G.S.() = 0.996 1 + 0.08962 and E.S.() = –0.996 1 + 0.08962b)G.S.() = 0.996 1 + 0.08962 and E.S.() = –0.0896 1 + 0.9962.c)G.S.() = –15.09 1 – 3.912 and E.S.() = –3.91 1 – 15.092d)G.S.() = 15.09 1 + 3.912 and E.S.() = –3.91 1 – 15.092Correct answer is option 'B'. 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 Assume that for a real system, a real wave function is a linear combination of two orthonormal basis functions where energy integrals are H11 = –15, H22 = –4, H12 = H21 = –1. What would be the ground state and excited wave functions?a)G.S.() = 0.996 1 + 0.08962 and E.S.() = –0.996 1 + 0.08962b)G.S.() = 0.996 1 + 0.08962 and E.S.() = –0.0896 1 + 0.9962.c)G.S.() = –15.09 1 – 3.912 and E.S.() = –3.91 1 – 15.092d)G.S.() = 15.09 1 + 3.912 and E.S.() = –3.91 1 – 15.092Correct answer is option 'B'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
Assume that for a real system, a real wave function is a linear combination of two orthonormal basis functions where energy integrals are H11 = –15, H22 = –4, H12 = H21 = –1. What would be the ground state and excited wave functions?a)G.S.() = 0.996 1 + 0.08962 and E.S.() = –0.996 1 + 0.08962b)G.S.() = 0.996 1 + 0.08962 and E.S.() = –0.0896 1 + 0.9962.c)G.S.() = –15.09 1 – 3.912 and E.S.() = –3.91 1 – 15.092d)G.S.() = 15.09 1 + 3.912 and E.S.() = –3.91 1 – 15.092Correct answer is option 'B'. Can you explain this answer?, a detailed solution for Assume that for a real system, a real wave function is a linear combination of two orthonormal basis functions where energy integrals are H11 = –15, H22 = –4, H12 = H21 = –1. What would be the ground state and excited wave functions?a)G.S.() = 0.996 1 + 0.08962 and E.S.() = –0.996 1 + 0.08962b)G.S.() = 0.996 1 + 0.08962 and E.S.() = –0.0896 1 + 0.9962.c)G.S.() = –15.09 1 – 3.912 and E.S.() = –3.91 1 – 15.092d)G.S.() = 15.09 1 + 3.912 and E.S.() = –3.91 1 – 15.092Correct answer is option 'B'. Can you explain this answer? has been provided alongside types of Assume that for a real system, a real wave function is a linear combination of two orthonormal basis functions where energy integrals are H11 = –15, H22 = –4, H12 = H21 = –1. What would be the ground state and excited wave functions?a)G.S.() = 0.996 1 + 0.08962 and E.S.() = –0.996 1 + 0.08962b)G.S.() = 0.996 1 + 0.08962 and E.S.() = –0.0896 1 + 0.9962.c)G.S.() = –15.09 1 – 3.912 and E.S.() = –3.91 1 – 15.092d)G.S.() = 15.09 1 + 3.912 and E.S.() = –3.91 1 – 15.092Correct answer is option 'B'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice Assume that for a real system, a real wave function is a linear combination of two orthonormal basis functions where energy integrals are H11 = –15, H22 = –4, H12 = H21 = –1. What would be the ground state and excited wave functions?a)G.S.() = 0.996 1 + 0.08962 and E.S.() = –0.996 1 + 0.08962b)G.S.() = 0.996 1 + 0.08962 and E.S.() = –0.0896 1 + 0.9962.c)G.S.() = –15.09 1 – 3.912 and E.S.() = –3.91 1 – 15.092d)G.S.() = 15.09 1 + 3.912 and E.S.() = –3.91 1 – 15.092Correct answer is option 'B'. 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