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If the shortest wavelength of lyman series of H atom is x, then the wavelength of first line of balmer series of H atom will be?
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If the shortest wavelength of lyman series of H atom is x, then the wa...
Calculation of Wavelength of First Line of Balmer Series of H Atom


The Lyman series of the hydrogen atom consists of transitions from higher energy levels to the lowest energy level (n=1), and the Balmer series consists of transitions from higher energy levels to the second energy level (n=2). Therefore, the Balmer series is the second spectral series of the hydrogen atom.


Formula for Calculation


The formula for calculating the wavelength of the Balmer series is given by:


1/λ = R(1/2^2 - 1/n^2)


Where λ is the wavelength of the spectral line, R is the Rydberg constant (1.0974 x 10^7 m^-1), and n is the principal quantum number of the final energy level.


For the first line of the Balmer series, the final energy level is n=2.


Calculation


Let the shortest wavelength of the Lyman series be x. The first line of the Balmer series corresponds to the transition from n=3 to n=2. Therefore, using the formula above:


1/λ = R(1/2^2 - 1/3^2)


1/λ = (1.0974 x 10^7 m^-1)(1/4 - 1/9)


1/λ = (1.0974 x 10^7 m^-1)(5/36)


1/λ = 1.2157 x 10^6 m^-1


λ = 8.225 x 10^-7 m, or 822.5 nm (rounded to three significant figures).


Conclusion


Therefore, the wavelength of the first line of the Balmer series of the hydrogen atom is 822.5 nm, assuming the shortest wavelength of the Lyman series is x.
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