UGC NET Exam > UGC NET Questions > Considering anharmonicity, the pattern of the...
Start Learning for Free
Considering anharmonicity, the pattern of the allowed vibrational energy levels may be given by:
For a molecule with ϖe = 4138.5 cm-1 and xe = 0.0218 the vibrational level of dissociation is
For a molecule with ϖe = 4138.5 cm-1 and xe = 0.0218 the vibrational level of dissociation is
- a)45
- b)22
- c)not determined, due to insufficient data
- d)90
Correct answer is option 'B'. Can you explain this answer?
| FREE This question is part of | Download PDF Attempt this Test |
Most Upvoted Answer
Considering anharmonicity, the pattern of the allowed vibrational ener...
The dissociation energy (De) in terms of wavenumbers for an anharmonic oscillator can be estimated using the first few terms of its vibrational energy level formula
E(v) = ωe(v + 1/2) - ωexe(v + 1/2)2, where v is the vibrational quantum number, ωe is the vibrational constant, and ωexe is the anharmonic correction to the vibrational constant.
For the dissociation energy,
De = ωe(vdiss + 1/2) - ωexe(vdiss + 1/2)2
We know that the energy at the point of dissociation (Ediss) equals the dissociation energy (De), so we can set E(vdiss) equal to De and solve for vdiss:
vdiss=(√(1/4∗(ωexe)2))+De/ωexe−1/2∗ωexe
Given ωe = 4138.5 cm-1 and ωexe = 0.0218, we can find the dissociation energy, assuming it's approximately equal to the potential well depth
(De = ωe/2ωexe):
De = ωe / (2 * ωexe) De = 4138.5 / (2 * 0.0218) De = 94,924.77 cm-1
Then, substituting ωe, ωexe and De into the v_diss equation we get:
vdiss ≈ 21.8
So, the vibrational level of dissociation is 22.
E(v) = ωe(v + 1/2) - ωexe(v + 1/2)2, where v is the vibrational quantum number, ωe is the vibrational constant, and ωexe is the anharmonic correction to the vibrational constant.
For the dissociation energy,
De = ωe(vdiss + 1/2) - ωexe(vdiss + 1/2)2
We know that the energy at the point of dissociation (Ediss) equals the dissociation energy (De), so we can set E(vdiss) equal to De and solve for vdiss:
vdiss=(√(1/4∗(ωexe)2))+De/ωexe−1/2∗ωexe
Given ωe = 4138.5 cm-1 and ωexe = 0.0218, we can find the dissociation energy, assuming it's approximately equal to the potential well depth
(De = ωe/2ωexe):
De = ωe / (2 * ωexe) De = 4138.5 / (2 * 0.0218) De = 94,924.77 cm-1
Then, substituting ωe, ωexe and De into the v_diss equation we get:
vdiss ≈ 21.8
So, the vibrational level of dissociation is 22.
Attention UGC NET Students!
To make sure you are not studying endlessly, EduRev has designed UGC NET study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in UGC NET.
|
Explore Courses for UGC NET exam
|
|
Top Courses for UGC NETView all
Considering anharmonicity, the pattern of the allowed vibrational energy levels may be given by:For a molecule withϖe = 4138.5 cm-1and xe= 0.0218 the vibrational level of dissociation isa)45b)22c)not determined, due to insufficient datad)90Correct answer is option 'B'. Can you explain this answer?
Question Description
Considering anharmonicity, the pattern of the allowed vibrational energy levels may be given by:For a molecule withϖe = 4138.5 cm-1and xe= 0.0218 the vibrational level of dissociation isa)45b)22c)not determined, due to insufficient datad)90Correct answer is option 'B'. Can you explain this answer? for UGC NET 2024 is part of UGC NET preparation. The Question and answers have been prepared according to the UGC NET exam syllabus. Information about Considering anharmonicity, the pattern of the allowed vibrational energy levels may be given by:For a molecule withϖe = 4138.5 cm-1and xe= 0.0218 the vibrational level of dissociation isa)45b)22c)not determined, due to insufficient datad)90Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for UGC NET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Considering anharmonicity, the pattern of the allowed vibrational energy levels may be given by:For a molecule withϖe = 4138.5 cm-1and xe= 0.0218 the vibrational level of dissociation isa)45b)22c)not determined, due to insufficient datad)90Correct answer is option 'B'. Can you explain this answer?.
Considering anharmonicity, the pattern of the allowed vibrational energy levels may be given by:For a molecule withϖe = 4138.5 cm-1and xe= 0.0218 the vibrational level of dissociation isa)45b)22c)not determined, due to insufficient datad)90Correct answer is option 'B'. Can you explain this answer? for UGC NET 2024 is part of UGC NET preparation. The Question and answers have been prepared according to the UGC NET exam syllabus. Information about Considering anharmonicity, the pattern of the allowed vibrational energy levels may be given by:For a molecule withϖe = 4138.5 cm-1and xe= 0.0218 the vibrational level of dissociation isa)45b)22c)not determined, due to insufficient datad)90Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for UGC NET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Considering anharmonicity, the pattern of the allowed vibrational energy levels may be given by:For a molecule withϖe = 4138.5 cm-1and xe= 0.0218 the vibrational level of dissociation isa)45b)22c)not determined, due to insufficient datad)90Correct answer is option 'B'. Can you explain this answer?.
Solutions for Considering anharmonicity, the pattern of the allowed vibrational energy levels may be given by:For a molecule withϖe = 4138.5 cm-1and xe= 0.0218 the vibrational level of dissociation isa)45b)22c)not determined, due to insufficient datad)90Correct answer is option 'B'. Can you explain this answer? in English & in Hindi are available as part of our courses for UGC NET.
Download more important topics, notes, lectures and mock test series for UGC NET Exam by signing up for free.
Here you can find the meaning of Considering anharmonicity, the pattern of the allowed vibrational energy levels may be given by:For a molecule withϖe = 4138.5 cm-1and xe= 0.0218 the vibrational level of dissociation isa)45b)22c)not determined, due to insufficient datad)90Correct answer is option 'B'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
Considering anharmonicity, the pattern of the allowed vibrational energy levels may be given by:For a molecule withϖe = 4138.5 cm-1and xe= 0.0218 the vibrational level of dissociation isa)45b)22c)not determined, due to insufficient datad)90Correct answer is option 'B'. Can you explain this answer?, a detailed solution for Considering anharmonicity, the pattern of the allowed vibrational energy levels may be given by:For a molecule withϖe = 4138.5 cm-1and xe= 0.0218 the vibrational level of dissociation isa)45b)22c)not determined, due to insufficient datad)90Correct answer is option 'B'. Can you explain this answer? has been provided alongside types of Considering anharmonicity, the pattern of the allowed vibrational energy levels may be given by:For a molecule withϖe = 4138.5 cm-1and xe= 0.0218 the vibrational level of dissociation isa)45b)22c)not determined, due to insufficient datad)90Correct answer is option 'B'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice Considering anharmonicity, the pattern of the allowed vibrational energy levels may be given by:For a molecule withϖe = 4138.5 cm-1and xe= 0.0218 the vibrational level of dissociation isa)45b)22c)not determined, due to insufficient datad)90Correct answer is option 'B'. Can you explain this answer? tests, examples and also practice UGC NET tests.
|
|
Explore Courses for UGC NET 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