The wavelength of the first line of Balmer series is 6563 A o . Find the wavelength of the first line of the Lyman series.

For the first line of the Lyman series n 11 = 1 and n 12 = 2 The wave number ν of the associated wave will be: ν = R H [ 1 n 2 11 − 1 n 2 12 ] cm − 1 Substitute the values in the equation, = 109677 [ 1 1 2 − 1 2 2 ] cm − 1 = 109677 × 3 4 cm − 1 ∵ λ = 1 ν λ = 4 3 × 1 109677 -------(i) For the first line of the Balmer series n 21 = 2 and n 22 = 3 The wave number ν of the associated wave will be: ν 2 = R H [ 1 n 2 21 − 1 n 2 22 ] cm − 1 Where, Rydberg Constant, R H = 109677 cm - 1 Substitute the values in the equation, ν 2 = 109677 [ 1 2 2 − 1 3 2 ] cm − 1 = 109677 × 5 36 cm − 1 ∵ λ 2 = 1 ν 2 = 36 5 × 1 109677 ---------(ii) Divide equation (ii) by equation (ii), We get : ∴ λ 2 λ = 36 5 4 3 We are given that the wavelength of balmer series = λ 2 = 6563 A ° Substituting this value in the above relation, we get: ⇒ 6563 λ = 27 5 ∴ λ = 1215 A °

The wavelength of the first member of the Balmer series is 6563 A o . Find the wavelengths of the second and the last member of the series.

For the first line of the Balmer series n 21 = 2 and n 22 = 3 The wave number ν of the associated wave will be: ν 1 = R H [ 1 n 2 21 − 1 n 2 22 ] cm − 1 Where, Rydberg Constant, R H = 109677 cm - 1 Substitute the values in the equation, = 109677 [ 1 2 2 − 1 3 2 ] cm − 1 = 109677 × 5 36 cm − 1 ∵ λ 1 = 1 ν 1 = 36 548385 ------(i) For the second line of the Balmer series n 21 = 2 and n 23 = 4 The wave number ν of the associated wave will be: ν 2 = R H [ 1 n 2 21 − 1 n 2 23 ] cm − 1 Where, Rydberg Constant, R H = 109677 cm - 1 Substitute the values in the equation, = 109677 [ 1 2 2 − 1 4 2 ] cm − 1 = 109677 × 3 16 cm − 1 ∵ λ 2 = 1 ν 2 = 16 329031 --------(ii) For the last line of the Balmer series the transition will take place between n 1 = 2 and n 2 = ∞ The wave number ν of the associated wave will be: ν ∞ = R H [ 1 n 2 1 − 1 n 2 2 ] cm − 1 Where, Rydberg Constant, R H = 109677 cm - 1 Substitute the values in the equation, = 109677 [ 1 2 2 − 1 ∞ 2 ] cm − 1 = 109677 × 1 4 cm − 1 ∵ λ ∞ = 1 ν = 4 109677 -----(iii) We are given λ 1 = 6563 A ° Divide equation (ii) by equation (i): λ 2 λ 1 = 20 27 λ 2 = 20 27 × 6563 = 4861 . 5 A ° Divide equation (iii) by equation (i): λ ∞ λ 1 = 20 36 λ ∞ = 20 36 × 6563 = 3646 . 1 A °

The wavelength of the first line of Balmer series is 6563 A o . Calculate Rydberg's constant.

For the first line of the Balmer series n 21 = 2 and n 22 = 3 We are given, the wavelength of first line of Balmer series = 6563 A ° = 6563 × 10 - 10 m The wave number ν of the associated wave will be: ν 1 = R H [ 1 n 2 21 − 1 n 2 22 ] Where, R H is the Rydberg Constant. Substitute the values in the equation, = R H [ 1 2 2 − 1 3 2 ] = R H × 5 36 ∵ λ 1 = 1 ν 1 ∴ 6563 × 10 - 10 m = 36 5 × R H ⇒ R H = 36 6563 × 10 - 10 m × 5 = 1 . 097 × 10 7 m - 1

E M B I B E

ISC Physics Class 12, Part-2>Chapter 13 - Atom, Origin of Spectra : Bohr's Theory of Hydrogen Atom>QUESTIONS>Q 28

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12th ICSE

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Calculate the wavelength of the first line of the Balmer series of hydrogen. Indicate the transition on an energy-level diagram. What is this line called?

Important Questions on Atom, Origin of Spectra : Bohr's Theory of Hydrogen Atom

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12th ICSE

IMPORTANT

ISC Physics Class 12, Part-2>Chapter 13 - Atom, Origin of Spectra : Bohr's Theory of Hydrogen Atom>QUESTIONS>Q 28

Calculate the wavelength of the first line of the Balmer series of hydrogen. Indicate the transition on an energy-level diagram. What is this line called? Does it lie in the visible region of the electro-magnetic spectrum?