IIT JAM Exam > IIT JAM Questions > In the hydrogen atom spectrum, the ratio of t...
Start Learning for Free
In the hydrogen atom spectrum, the ratio of the longest wavelength in the Lyman series (final state n = 1) to that in the Balmer series (final state n = 2) is ___________.
Correct answer is '0.185'. Can you explain this answer?
| FREE This question is part of | Download PDF Attempt this Test |
Most Upvoted Answer
In the hydrogen atom spectrum, the ratio of the longest wavelength in ...
Ratio of Longest Wavelengths in Lyman and Balmer Series in Hydrogen Atom Spectrum
The hydrogen atom spectrum consists of a series of spectral lines that are produced when an electron transitions between different energy levels within the atom. These spectral lines are characterized by their wavelengths, and they can be grouped into different series based on the final energy level of the electron.
The Lyman series corresponds to transitions where the electron's final energy level is n = 1, while the Balmer series corresponds to transitions where the final energy level is n = 2. The longest wavelength in each series corresponds to the lowest energy transition within that series.
Lyman Series:
In the Lyman series, the electron transitions from higher energy levels (n > 1) to the lowest energy level (n = 1). The formula for calculating the wavelength of a spectral line in the Lyman series is given by:
1/λ = R_H * (1 - 1/n^2)
Where λ is the wavelength, R_H is the Rydberg constant for hydrogen (approximately 1.097 x 10^7 m^-1), and n is the energy level of the electron.
For the longest wavelength in the Lyman series, we need to find the transition with the smallest change in energy level. This occurs when n = 2. Plugging in n = 2 into the formula, we get:
1/λ_Lyman = R_H * (1 - 1/2^2)
1/λ_Lyman = R_H * (1 - 1/4)
1/λ_Lyman = R_H * (3/4)
Balmer Series:
In the Balmer series, the electron transitions from higher energy levels (n > 2) to the second energy level (n = 2). The formula for calculating the wavelength of a spectral line in the Balmer series is similar to the Lyman series:
1/λ = R_H * (1 - 1/n^2)
For the longest wavelength in the Balmer series, we need to find the transition with the smallest change in energy level. This occurs when n = 3. Plugging in n = 3 into the formula, we get:
1/λ_Balmer = R_H * (1 - 1/3^2)
1/λ_Balmer = R_H * (1 - 1/9)
1/λ_Balmer = R_H * (8/9)
Ratio of Longest Wavelengths:
To find the ratio of the longest wavelengths in the Lyman and Balmer series, we divide the equation for the Lyman series by the equation for the Balmer series:
(1/λ_Lyman) / (1/λ_Balmer) = (R_H * (3/4)) / (R_H * (8/9))
(1/λ_Lyman) / (1/λ_Balmer) = (3/4) / (8/9)
(1/λ_Lyman) / (1/λ_Balmer) = (3/4) * (9/8)
(1/λ_Lyman) / (1/λ_Balmer) = 27/32
So, the ratio of the longest wavelength in the Lyman series to
The hydrogen atom spectrum consists of a series of spectral lines that are produced when an electron transitions between different energy levels within the atom. These spectral lines are characterized by their wavelengths, and they can be grouped into different series based on the final energy level of the electron.
The Lyman series corresponds to transitions where the electron's final energy level is n = 1, while the Balmer series corresponds to transitions where the final energy level is n = 2. The longest wavelength in each series corresponds to the lowest energy transition within that series.
Lyman Series:
In the Lyman series, the electron transitions from higher energy levels (n > 1) to the lowest energy level (n = 1). The formula for calculating the wavelength of a spectral line in the Lyman series is given by:
1/λ = R_H * (1 - 1/n^2)
Where λ is the wavelength, R_H is the Rydberg constant for hydrogen (approximately 1.097 x 10^7 m^-1), and n is the energy level of the electron.
For the longest wavelength in the Lyman series, we need to find the transition with the smallest change in energy level. This occurs when n = 2. Plugging in n = 2 into the formula, we get:
1/λ_Lyman = R_H * (1 - 1/2^2)
1/λ_Lyman = R_H * (1 - 1/4)
1/λ_Lyman = R_H * (3/4)
Balmer Series:
In the Balmer series, the electron transitions from higher energy levels (n > 2) to the second energy level (n = 2). The formula for calculating the wavelength of a spectral line in the Balmer series is similar to the Lyman series:
1/λ = R_H * (1 - 1/n^2)
For the longest wavelength in the Balmer series, we need to find the transition with the smallest change in energy level. This occurs when n = 3. Plugging in n = 3 into the formula, we get:
1/λ_Balmer = R_H * (1 - 1/3^2)
1/λ_Balmer = R_H * (1 - 1/9)
1/λ_Balmer = R_H * (8/9)
Ratio of Longest Wavelengths:
To find the ratio of the longest wavelengths in the Lyman and Balmer series, we divide the equation for the Lyman series by the equation for the Balmer series:
(1/λ_Lyman) / (1/λ_Balmer) = (R_H * (3/4)) / (R_H * (8/9))
(1/λ_Lyman) / (1/λ_Balmer) = (3/4) / (8/9)
(1/λ_Lyman) / (1/λ_Balmer) = (3/4) * (9/8)
(1/λ_Lyman) / (1/λ_Balmer) = 27/32
So, the ratio of the longest wavelength in the Lyman series to
|
Explore Courses for IIT JAM exam
|
|
Similar IIT JAM Doubts
In the hydrogen atom spectrum, the ratio of the longest wavelength in the Lyman series (final state n = 1) to that in the Balmer series (final state n = 2) is ___________.Correct answer is '0.185'. Can you explain this answer?
Question Description
In the hydrogen atom spectrum, the ratio of the longest wavelength in the Lyman series (final state n = 1) to that in the Balmer series (final state n = 2) is ___________.Correct answer is '0.185'. Can you explain this answer? for IIT JAM 2024 is part of IIT JAM preparation. The Question and answers have been prepared according to the IIT JAM exam syllabus. Information about In the hydrogen atom spectrum, the ratio of the longest wavelength in the Lyman series (final state n = 1) to that in the Balmer series (final state n = 2) is ___________.Correct answer is '0.185'. Can you explain this answer? covers all topics & solutions for IIT JAM 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In the hydrogen atom spectrum, the ratio of the longest wavelength in the Lyman series (final state n = 1) to that in the Balmer series (final state n = 2) is ___________.Correct answer is '0.185'. Can you explain this answer?.
In the hydrogen atom spectrum, the ratio of the longest wavelength in the Lyman series (final state n = 1) to that in the Balmer series (final state n = 2) is ___________.Correct answer is '0.185'. Can you explain this answer? for IIT JAM 2024 is part of IIT JAM preparation. The Question and answers have been prepared according to the IIT JAM exam syllabus. Information about In the hydrogen atom spectrum, the ratio of the longest wavelength in the Lyman series (final state n = 1) to that in the Balmer series (final state n = 2) is ___________.Correct answer is '0.185'. Can you explain this answer? covers all topics & solutions for IIT JAM 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In the hydrogen atom spectrum, the ratio of the longest wavelength in the Lyman series (final state n = 1) to that in the Balmer series (final state n = 2) is ___________.Correct answer is '0.185'. Can you explain this answer?.
Solutions for In the hydrogen atom spectrum, the ratio of the longest wavelength in the Lyman series (final state n = 1) to that in the Balmer series (final state n = 2) is ___________.Correct answer is '0.185'. Can you explain this answer? in English & in Hindi are available as part of our courses for IIT JAM.
Download more important topics, notes, lectures and mock test series for IIT JAM Exam by signing up for free.
Here you can find the meaning of In the hydrogen atom spectrum, the ratio of the longest wavelength in the Lyman series (final state n = 1) to that in the Balmer series (final state n = 2) is ___________.Correct answer is '0.185'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
In the hydrogen atom spectrum, the ratio of the longest wavelength in the Lyman series (final state n = 1) to that in the Balmer series (final state n = 2) is ___________.Correct answer is '0.185'. Can you explain this answer?, a detailed solution for In the hydrogen atom spectrum, the ratio of the longest wavelength in the Lyman series (final state n = 1) to that in the Balmer series (final state n = 2) is ___________.Correct answer is '0.185'. Can you explain this answer? has been provided alongside types of In the hydrogen atom spectrum, the ratio of the longest wavelength in the Lyman series (final state n = 1) to that in the Balmer series (final state n = 2) is ___________.Correct answer is '0.185'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice In the hydrogen atom spectrum, the ratio of the longest wavelength in the Lyman series (final state n = 1) to that in the Balmer series (final state n = 2) is ___________.Correct answer is '0.185'. Can you explain this answer? tests, examples and also practice IIT JAM tests.
|
|
Explore Courses for IIT JAM 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