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One of the lines in the emission spectrum of Li2+ has the same wavelength as that of the 2nd line of Balmer series in hydrogen spectrum. The electronic transition corresponding to this line is n = 12 → n = x. Find the value of x.
- a)8
- b)6
- c)7
- d)5
Correct answer is option 'B'. Can you explain this answer?
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One of the lines in the emission spectrum of Li2+ has the same wavelen...
For 2nd line of Balmer series in hydrogen spectrum 
which is satisfied by n = 12 → n = 6.
which is satisfied by n = 12 → n = 6.
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One of the lines in the emission spectrum of Li2+ has the same wavelen...
To determine the electronic transition corresponding to the line in the emission spectrum of Li2, we need to compare it to the Balmer series of hydrogen.
The Balmer series in the hydrogen spectrum is given by the formula:
1/λ = R(1/n1^2 - 1/n2^2)
Where λ is the wavelength, R is the Rydberg constant (approximately 1.097 × 10^7 m^-1), n1 is the initial energy level, and n2 is the final energy level.
The 2nd line of the Balmer series corresponds to n1 = 2 and n2 = 3. So,
1/λ_hydrogen = R(1/2^2 - 1/3^2)
1/λ_hydrogen = R(1/4 - 1/9)
1/λ_hydrogen = R(5/36)
Now, we know that the line in the emission spectrum of Li2 has the same wavelength as the 2nd line of the Balmer series in hydrogen. So,
1/λ_li2 = 1/λ_hydrogen
1/λ_li2 = R(5/36)
To find the electronic transition in Li2, we need to determine the value of n for this transition. Rearranging the equation above, we have:
λ_li2 = 36/(5R)
Since R is a constant, the value of λ_li2 will depend on the value of n. We need to find the value of n such that the wavelength matches λ_li2.
Let's assume the electronic transition in Li2 corresponds to n = 12. Plugging this value into the equation, we get:
λ_li2 = 36/(5R)
λ_li2 = 36/(5 * 1.097 × 10^7)
λ_li2 ≈ 6.53 × 10^-8 m
Therefore, if the electronic transition in Li2 corresponds to n = 12, the wavelength of the line in its emission spectrum will be approximately 6.53 × 10^-8 meters.
The Balmer series in the hydrogen spectrum is given by the formula:
1/λ = R(1/n1^2 - 1/n2^2)
Where λ is the wavelength, R is the Rydberg constant (approximately 1.097 × 10^7 m^-1), n1 is the initial energy level, and n2 is the final energy level.
The 2nd line of the Balmer series corresponds to n1 = 2 and n2 = 3. So,
1/λ_hydrogen = R(1/2^2 - 1/3^2)
1/λ_hydrogen = R(1/4 - 1/9)
1/λ_hydrogen = R(5/36)
Now, we know that the line in the emission spectrum of Li2 has the same wavelength as the 2nd line of the Balmer series in hydrogen. So,
1/λ_li2 = 1/λ_hydrogen
1/λ_li2 = R(5/36)
To find the electronic transition in Li2, we need to determine the value of n for this transition. Rearranging the equation above, we have:
λ_li2 = 36/(5R)
Since R is a constant, the value of λ_li2 will depend on the value of n. We need to find the value of n such that the wavelength matches λ_li2.
Let's assume the electronic transition in Li2 corresponds to n = 12. Plugging this value into the equation, we get:
λ_li2 = 36/(5R)
λ_li2 = 36/(5 * 1.097 × 10^7)
λ_li2 ≈ 6.53 × 10^-8 m
Therefore, if the electronic transition in Li2 corresponds to n = 12, the wavelength of the line in its emission spectrum will be approximately 6.53 × 10^-8 meters.
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One of the lines in the emission spectrum of Li2+ has the same wavelength as that of the 2nd line of Balmer series in hydrogen spectrum. The electronic transition corresponding to this line is n = 12 → n = x. Find the value of x.a)8b)6c)7d)5Correct answer is option 'B'. Can you explain this answer?
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One of the lines in the emission spectrum of Li2+ has the same wavelength as that of the 2nd line of Balmer series in hydrogen spectrum. The electronic transition corresponding to this line is n = 12 → n = x. Find the value of x.a)8b)6c)7d)5Correct answer is option 'B'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about One of the lines in the emission spectrum of Li2+ has the same wavelength as that of the 2nd line of Balmer series in hydrogen spectrum. The electronic transition corresponding to this line is n = 12 → n = x. Find the value of x.a)8b)6c)7d)5Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for One of the lines in the emission spectrum of Li2+ has the same wavelength as that of the 2nd line of Balmer series in hydrogen spectrum. The electronic transition corresponding to this line is n = 12 → n = x. Find the value of x.a)8b)6c)7d)5Correct answer is option 'B'. Can you explain this answer?.
One of the lines in the emission spectrum of Li2+ has the same wavelength as that of the 2nd line of Balmer series in hydrogen spectrum. The electronic transition corresponding to this line is n = 12 → n = x. Find the value of x.a)8b)6c)7d)5Correct answer is option 'B'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about One of the lines in the emission spectrum of Li2+ has the same wavelength as that of the 2nd line of Balmer series in hydrogen spectrum. The electronic transition corresponding to this line is n = 12 → n = x. Find the value of x.a)8b)6c)7d)5Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for One of the lines in the emission spectrum of Li2+ has the same wavelength as that of the 2nd line of Balmer series in hydrogen spectrum. The electronic transition corresponding to this line is n = 12 → n = x. Find the value of x.a)8b)6c)7d)5Correct answer is option 'B'. Can you explain this answer?.
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One of the lines in the emission spectrum of Li2+ has the same wavelength as that of the 2nd line of Balmer series in hydrogen spectrum. The electronic transition corresponding to this line is n = 12 → n = x. Find the value of x.a)8b)6c)7d)5Correct answer is option 'B'. Can you explain this answer?, a detailed solution for One of the lines in the emission spectrum of Li2+ has the same wavelength as that of the 2nd line of Balmer series in hydrogen spectrum. The electronic transition corresponding to this line is n = 12 → n = x. Find the value of x.a)8b)6c)7d)5Correct answer is option 'B'. Can you explain this answer? has been provided alongside types of One of the lines in the emission spectrum of Li2+ has the same wavelength as that of the 2nd line of Balmer series in hydrogen spectrum. The electronic transition corresponding to this line is n = 12 → n = x. Find the value of x.a)8b)6c)7d)5Correct answer is option 'B'. Can you explain this answer? theory, EduRev gives you an
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