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The ratio of difference in wavelengths of 1st and 2nd lines of Lyman series in H-like atom to difference in wavelength for 2nd and 3rd lines of same series is :
- a)2.5 : 1
- b)3.5 : 1
- c)4.5 : 1
- d)5.5 : 1
Correct answer is option 'B'. Can you explain this answer?
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The ratio of difference in wavelengths of 1st and 2nd lines of Lyman s...
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The ratio of difference in wavelengths of 1st and 2nd lines of Lyman s...
Explanation:
The Lyman series is a series of spectral lines that appear in the ultraviolet region of the electromagnetic spectrum when an electron in a hydrogen-like atom transitions from a higher energy level to the first energy level (n = 1). The series is named after Theodore Lyman, who first observed these lines.
Wavelengths of the Lyman Series:
The wavelengths of the spectral lines in the Lyman series can be calculated using the Rydberg formula:
1/λ = R(1 - 1/n^2)
Where:
- λ is the wavelength of the spectral line
- R is the Rydberg constant (approximately 1.097 x 10^7 m^-1)
- n is the principal quantum number of the energy level
We are given that we need to find the ratio of the difference in wavelengths between the 1st and 2nd lines of the Lyman series to the difference in wavelengths between the 2nd and 3rd lines of the same series.
Calculating the Wavelengths:
For the 1st line of the Lyman series, n1 = 2. Plugging this value into the Rydberg formula:
1/λ1 = R(1 - 1/2^2)
1/λ1 = R(1 - 1/4)
1/λ1 = R(3/4)
λ1 = 4/3R
For the 2nd line of the Lyman series, n2 = 3. Plugging this value into the Rydberg formula:
1/λ2 = R(1 - 1/3^2)
1/λ2 = R(1 - 1/9)
1/λ2 = R(8/9)
λ2 = 9/8R
For the 3rd line of the Lyman series, n3 = 4. Plugging this value into the Rydberg formula:
1/λ3 = R(1 - 1/4^2)
1/λ3 = R(1 - 1/16)
1/λ3 = R(15/16)
λ3 = 16/15R
Calculating the Ratios:
The difference in wavelengths between the 1st and 2nd lines is given by:
Δλ1-2 = λ2 - λ1
Δλ1-2 = (9/8R) - (4/3R)
Δλ1-2 = (27R - 32R) / (24R)
Δλ1-2 = -5R / (24R)
Δλ1-2 = -5/24
The difference in wavelengths between the 2nd and 3rd lines is given by:
Δλ2-3 = λ3 - λ2
Δλ2-3 = (16/15R) - (9/8R)
Δλ2-3 = (128R - 135R) / (120R)
Δλ2-3 = -7R / (120R)
Δλ2-3 = -7/120
Taking the ratio of the two differences:
Ratio = Δλ1-2 / Δλ2-3
Ratio = (-5/24) / (-7/120
The Lyman series is a series of spectral lines that appear in the ultraviolet region of the electromagnetic spectrum when an electron in a hydrogen-like atom transitions from a higher energy level to the first energy level (n = 1). The series is named after Theodore Lyman, who first observed these lines.
Wavelengths of the Lyman Series:
The wavelengths of the spectral lines in the Lyman series can be calculated using the Rydberg formula:
1/λ = R(1 - 1/n^2)
Where:
- λ is the wavelength of the spectral line
- R is the Rydberg constant (approximately 1.097 x 10^7 m^-1)
- n is the principal quantum number of the energy level
We are given that we need to find the ratio of the difference in wavelengths between the 1st and 2nd lines of the Lyman series to the difference in wavelengths between the 2nd and 3rd lines of the same series.
Calculating the Wavelengths:
For the 1st line of the Lyman series, n1 = 2. Plugging this value into the Rydberg formula:
1/λ1 = R(1 - 1/2^2)
1/λ1 = R(1 - 1/4)
1/λ1 = R(3/4)
λ1 = 4/3R
For the 2nd line of the Lyman series, n2 = 3. Plugging this value into the Rydberg formula:
1/λ2 = R(1 - 1/3^2)
1/λ2 = R(1 - 1/9)
1/λ2 = R(8/9)
λ2 = 9/8R
For the 3rd line of the Lyman series, n3 = 4. Plugging this value into the Rydberg formula:
1/λ3 = R(1 - 1/4^2)
1/λ3 = R(1 - 1/16)
1/λ3 = R(15/16)
λ3 = 16/15R
Calculating the Ratios:
The difference in wavelengths between the 1st and 2nd lines is given by:
Δλ1-2 = λ2 - λ1
Δλ1-2 = (9/8R) - (4/3R)
Δλ1-2 = (27R - 32R) / (24R)
Δλ1-2 = -5R / (24R)
Δλ1-2 = -5/24
The difference in wavelengths between the 2nd and 3rd lines is given by:
Δλ2-3 = λ3 - λ2
Δλ2-3 = (16/15R) - (9/8R)
Δλ2-3 = (128R - 135R) / (120R)
Δλ2-3 = -7R / (120R)
Δλ2-3 = -7/120
Taking the ratio of the two differences:
Ratio = Δλ1-2 / Δλ2-3
Ratio = (-5/24) / (-7/120
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The ratio of difference in wavelengths of 1st and 2nd lines of Lyman s...
B
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The ratio of difference in wavelengths of 1st and 2nd lines of Lyman series in H-like atom to difference in wavelength for 2nd and 3rd lines of same series is :a)2.5 : 1b)3.5 : 1c)4.5 : 1d)5.5 : 1Correct answer is option 'B'. Can you explain this answer?
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The ratio of difference in wavelengths of 1st and 2nd lines of Lyman series in H-like atom to difference in wavelength for 2nd and 3rd lines of same series is :a)2.5 : 1b)3.5 : 1c)4.5 : 1d)5.5 : 1Correct 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 ratio of difference in wavelengths of 1st and 2nd lines of Lyman series in H-like atom to difference in wavelength for 2nd and 3rd lines of same series is :a)2.5 : 1b)3.5 : 1c)4.5 : 1d)5.5 : 1Correct 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 ratio of difference in wavelengths of 1st and 2nd lines of Lyman series in H-like atom to difference in wavelength for 2nd and 3rd lines of same series is :a)2.5 : 1b)3.5 : 1c)4.5 : 1d)5.5 : 1Correct answer is option 'B'. Can you explain this answer?.
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