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Find the longest wavelength present in the Balmer series of hydrogen, corresponding to the H- line.
- a)656 nm
- b)676 nm
- c)666 nm
- d)686 nm
Correct answer is option 'A'. Can you explain this answer?
Most Upvoted Answer
Find the longest wavelength present in the Balmer series of hydrogen, ...
If that is so grant us some suggestion! Or Balmer sequence (2nd sequence from electrons laying off to n = 2 from bigger than 2) longest wavelength [velocityconstant = frequency x wavelength] is smallest frequency [power = hconstant x frequency] is least enegy is from transition n = 3 to n = 2 or Google it 656.3 nm
Community Answer
Find the longest wavelength present in the Balmer series of hydrogen, ...
Explanation:
The Balmer series of hydrogen is a set of spectral lines in the visible region of the electromagnetic spectrum. These lines are caused by the transition of electrons from higher energy levels to the second energy level of hydrogen atoms. The longest wavelength present in the Balmer series corresponds to the H-alpha line.
Formula:
The Balmer series can be represented by the formula:
1/λ = R [1/22 - 1/n2]
where λ is the wavelength of the spectral line, R is the Rydberg constant (1.097 x 107 m-1), and n is an integer representing the energy level of the electron.
Solving the equation for the H-alpha line (n = 3, representing the transition of an electron from the third energy level to the second energy level), we get:
1/λ = R [1/22 - 1/32]
1/λ = R (1/4 - 1/9)
1/λ = R (5/36)
λ = 36/(5R)
λ = 656.3 nm
Therefore, the longest wavelength present in the Balmer series of hydrogen, corresponding to the H-alpha line, is 656 nm, which is option A.
The Balmer series of hydrogen is a set of spectral lines in the visible region of the electromagnetic spectrum. These lines are caused by the transition of electrons from higher energy levels to the second energy level of hydrogen atoms. The longest wavelength present in the Balmer series corresponds to the H-alpha line.
Formula:
The Balmer series can be represented by the formula:
1/λ = R [1/22 - 1/n2]
where λ is the wavelength of the spectral line, R is the Rydberg constant (1.097 x 107 m-1), and n is an integer representing the energy level of the electron.
Solving the equation for the H-alpha line (n = 3, representing the transition of an electron from the third energy level to the second energy level), we get:
1/λ = R [1/22 - 1/32]
1/λ = R (1/4 - 1/9)
1/λ = R (5/36)
λ = 36/(5R)
λ = 656.3 nm
Therefore, the longest wavelength present in the Balmer series of hydrogen, corresponding to the H-alpha line, is 656 nm, which is option A.
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