Class 11 Exam > Class 11 Questions > The shortest wavelength of He atom in Balmer ...
Start Learning for Free
The shortest wavelength of He atom in Balmer series is x, then longest wavelength in the Paschene series of Li+2 is :
- a)36x/5
- b)16x/7
- c)9x/5
- d)5x/9
Correct answer is option 'B'. Can you explain this answer?
| FREE This question is part of | Download PDF Attempt this Test |
Verified Answer
The shortest wavelength of He atom in Balmer series is x, then longest...
The shortest wavelength of He+ ion spectrum in Balmer series is the transition from n = infinity ----> n = 2.
Hence, 1/λ1 = 1/x = RZ^2(1/2^2 − 1/∞^2)= R 2^2(1/4) = R
or R = 1/x
Then, the longest wavelength in the Paschen series of line of Li2+ ion will be from n = 4 -----> n = 3
1/λ2 = RZ^2(1/3^2 − 1/4^2) = R 3^2(1/9 − 1/16) = R 9(7/144)=R 7/16
or Since, R = 1/x
1/λ2 = 1/x x 7/16
Or λ2 =16x/7
Most Upvoted Answer
The shortest wavelength of He atom in Balmer series is x, then longest...
Shortest wavelength of He atom in Balmer series
The Balmer series of hydrogen-like atoms consists of spectral lines that result from electron transitions from energy levels n ≥ 3 to n = 2. For He atom, the Balmer series corresponds to n = 3 to n = 2 transitions.
Let λ be the shortest wavelength in the Balmer series of He atom. Then, we have:
1/λ = R(1/2^2 - 1/3^2)
where R is the Rydberg constant for He atom.
Simplifying the above equation, we get:
λ = 8R/5
Longest wavelength in the Paschen series of Li 2
The Paschen series of hydrogen-like atoms consists of spectral lines that result from electron transitions from energy levels n ≥ 4 to n = 3. For Li 2, the Paschen series corresponds to n = 4 to n = 3 transitions.
Let λ' be the longest wavelength in the Paschen series of Li 2. Then, we have:
1/λ' = R'(1/3^2 - 1/4^2)
where R' is the Rydberg constant for Li 2.
Simplifying the above equation, we get:
λ' = 16R'/7
Relation between Rydberg constants for He and Li 2
The Rydberg constant for hydrogen-like atoms is given by:
R∝m/e^2
where m is the reduced mass of the atom and e is the charge on the electron.
For He atom, we have:
m = 2m_e
where m_e is the mass of electron.
For Li 2, we have:
m = m_e/2
Substituting these values in the above equation, we get:
R/R' = 4
Using this relation, we can write:
λ' = 16R'/7 = 16R/(4*7) = 4λ/7
Substituting the value of λ from the first part, we get:
λ' = 16x/7
Therefore, the correct option is B.
The Balmer series of hydrogen-like atoms consists of spectral lines that result from electron transitions from energy levels n ≥ 3 to n = 2. For He atom, the Balmer series corresponds to n = 3 to n = 2 transitions.
Let λ be the shortest wavelength in the Balmer series of He atom. Then, we have:
1/λ = R(1/2^2 - 1/3^2)
where R is the Rydberg constant for He atom.
Simplifying the above equation, we get:
λ = 8R/5
Longest wavelength in the Paschen series of Li 2
The Paschen series of hydrogen-like atoms consists of spectral lines that result from electron transitions from energy levels n ≥ 4 to n = 3. For Li 2, the Paschen series corresponds to n = 4 to n = 3 transitions.
Let λ' be the longest wavelength in the Paschen series of Li 2. Then, we have:
1/λ' = R'(1/3^2 - 1/4^2)
where R' is the Rydberg constant for Li 2.
Simplifying the above equation, we get:
λ' = 16R'/7
Relation between Rydberg constants for He and Li 2
The Rydberg constant for hydrogen-like atoms is given by:
R∝m/e^2
where m is the reduced mass of the atom and e is the charge on the electron.
For He atom, we have:
m = 2m_e
where m_e is the mass of electron.
For Li 2, we have:
m = m_e/2
Substituting these values in the above equation, we get:
R/R' = 4
Using this relation, we can write:
λ' = 16R'/7 = 16R/(4*7) = 4λ/7
Substituting the value of λ from the first part, we get:
λ' = 16x/7
Therefore, the correct option is B.
Attention Class 11 Students!
To make sure you are not studying endlessly, EduRev has designed Class 11 study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in Class 11.
|
Explore Courses for Class 11 exam
|
|
Similar Class 11 Doubts
Top Courses for Class 11View all
The shortest wavelength of He atom in Balmer series is x, then longest wavelength in the Paschene series of Li+2is :a)36x/5b)16x/7c)9x/5d)5x/9Correct answer is option 'B'. Can you explain this answer?
Question Description
The shortest wavelength of He atom in Balmer series is x, then longest wavelength in the Paschene series of Li+2is :a)36x/5b)16x/7c)9x/5d)5x/9Correct answer is option 'B'. Can you explain this answer? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about The shortest wavelength of He atom in Balmer series is x, then longest wavelength in the Paschene series of Li+2is :a)36x/5b)16x/7c)9x/5d)5x/9Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The shortest wavelength of He atom in Balmer series is x, then longest wavelength in the Paschene series of Li+2is :a)36x/5b)16x/7c)9x/5d)5x/9Correct answer is option 'B'. Can you explain this answer?.
The shortest wavelength of He atom in Balmer series is x, then longest wavelength in the Paschene series of Li+2is :a)36x/5b)16x/7c)9x/5d)5x/9Correct answer is option 'B'. Can you explain this answer? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about The shortest wavelength of He atom in Balmer series is x, then longest wavelength in the Paschene series of Li+2is :a)36x/5b)16x/7c)9x/5d)5x/9Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The shortest wavelength of He atom in Balmer series is x, then longest wavelength in the Paschene series of Li+2is :a)36x/5b)16x/7c)9x/5d)5x/9Correct answer is option 'B'. Can you explain this answer?.
Solutions for The shortest wavelength of He atom in Balmer series is x, then longest wavelength in the Paschene series of Li+2is :a)36x/5b)16x/7c)9x/5d)5x/9Correct answer is option 'B'. Can you explain this answer? in English & in Hindi are available as part of our courses for Class 11.
Download more important topics, notes, lectures and mock test series for Class 11 Exam by signing up for free.
Here you can find the meaning of The shortest wavelength of He atom in Balmer series is x, then longest wavelength in the Paschene series of Li+2is :a)36x/5b)16x/7c)9x/5d)5x/9Correct answer is option 'B'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
The shortest wavelength of He atom in Balmer series is x, then longest wavelength in the Paschene series of Li+2is :a)36x/5b)16x/7c)9x/5d)5x/9Correct answer is option 'B'. Can you explain this answer?, a detailed solution for The shortest wavelength of He atom in Balmer series is x, then longest wavelength in the Paschene series of Li+2is :a)36x/5b)16x/7c)9x/5d)5x/9Correct answer is option 'B'. Can you explain this answer? has been provided alongside types of The shortest wavelength of He atom in Balmer series is x, then longest wavelength in the Paschene series of Li+2is :a)36x/5b)16x/7c)9x/5d)5x/9Correct answer is option 'B'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice The shortest wavelength of He atom in Balmer series is x, then longest wavelength in the Paschene series of Li+2is :a)36x/5b)16x/7c)9x/5d)5x/9Correct answer is option 'B'. Can you explain this answer? tests, examples and also practice Class 11 tests.
|
|
Explore Courses for Class 11 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
|
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup