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Taking the wavelength of first Balmer line in hydrogen spectrum (n = 3 to n = 2) as 660 nm, the wavelength of the 2nd Balmer line (n = 4 to n = 2) will be:
- a)642.7 nm
- b)889.2 nm
- c)488.9 nm
- d)388.9 nm
Correct answer is option 'C'. Can you explain this answer?
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Taking the wavelength of first Balmer line in hydrogen spectrum (n = 3...
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Taking the wavelength of first Balmer line in hydrogen spectrum (n = 3...
The Balmer series is a series of spectral lines in the hydrogen spectrum that corresponds to electron transitions from higher energy levels (n > 2) to the second energy level (n = 2). The wavelengths of the lines in the Balmer series can be calculated using the Balmer formula:
1/λ = R * (1/2^2 - 1/n^2)
where λ is the wavelength of the line, R is the Rydberg constant (approximately 1.097 * 10^7 m^-1), and n is the principal quantum number.
The given question states that the wavelength of the first Balmer line (n = 3 to n = 2) is 660 nm. We can use this information to find the value of R:
1/λ = R * (1/2^2 - 1/3^2)
1/660 * 10^-9 = R * (1/4 - 1/9)
R = 1/660 * 10^-9 / (1/4 - 1/9)
R ≈ 1.097 * 10^7 m^-1
Now, we can use the value of R to calculate the wavelength of the second Balmer line (n = 4 to n = 2). Plugging in the values into the Balmer formula:
1/λ = R * (1/2^2 - 1/4^2)
1/λ = R * (1/4 - 1/16)
1/λ = R * (3/16)
λ = 16/3R
λ ≈ 16/3 * 1.097 * 10^7
λ ≈ 5.872 * 10^-7 m
Converting the wavelength to nanometers:
λ ≈ 587.2 nm
Therefore, the correct answer is option (c) 488.9 nm.
1/λ = R * (1/2^2 - 1/n^2)
where λ is the wavelength of the line, R is the Rydberg constant (approximately 1.097 * 10^7 m^-1), and n is the principal quantum number.
The given question states that the wavelength of the first Balmer line (n = 3 to n = 2) is 660 nm. We can use this information to find the value of R:
1/λ = R * (1/2^2 - 1/3^2)
1/660 * 10^-9 = R * (1/4 - 1/9)
R = 1/660 * 10^-9 / (1/4 - 1/9)
R ≈ 1.097 * 10^7 m^-1
Now, we can use the value of R to calculate the wavelength of the second Balmer line (n = 4 to n = 2). Plugging in the values into the Balmer formula:
1/λ = R * (1/2^2 - 1/4^2)
1/λ = R * (1/4 - 1/16)
1/λ = R * (3/16)
λ = 16/3R
λ ≈ 16/3 * 1.097 * 10^7
λ ≈ 5.872 * 10^-7 m
Converting the wavelength to nanometers:
λ ≈ 587.2 nm
Therefore, the correct answer is option (c) 488.9 nm.
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Taking the wavelength of first Balmer line in hydrogen spectrum (n = 3 to n = 2) as 660 nm, the wavelength of the 2nd Balmer line (n = 4 to n = 2) will be:a)642.7 nmb)889.2 nmc)488.9 nmd)388.9 nmCorrect answer is option 'C'. Can you explain this answer?
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Taking the wavelength of first Balmer line in hydrogen spectrum (n = 3 to n = 2) as 660 nm, the wavelength of the 2nd Balmer line (n = 4 to n = 2) will be:a)642.7 nmb)889.2 nmc)488.9 nmd)388.9 nmCorrect answer is option 'C'. 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 Taking the wavelength of first Balmer line in hydrogen spectrum (n = 3 to n = 2) as 660 nm, the wavelength of the 2nd Balmer line (n = 4 to n = 2) will be:a)642.7 nmb)889.2 nmc)488.9 nmd)388.9 nmCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Taking the wavelength of first Balmer line in hydrogen spectrum (n = 3 to n = 2) as 660 nm, the wavelength of the 2nd Balmer line (n = 4 to n = 2) will be:a)642.7 nmb)889.2 nmc)488.9 nmd)388.9 nmCorrect answer is option 'C'. Can you explain this answer?.
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