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 :

Engineering

Physics

Spectrum of Hydrogen and Hydrogen Like Atoms

Bohrs Model of Atom

Question

Taking the wavelength of first Balmer line in hydrogen spectrum (n = 3 to n = 2) as 660 nm, the wavelength of the 2^nd Balmer line (n = 4 to n = 2) will be :

388.9 nm

642.7 nm

488.9 nm

889.2 nm

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$\frac{1}{660} = R (\frac{1}{2^{2}} - \frac{1}{3^{2}}) = \frac{5 R}{36}$ ....(1)

$\frac{1}{λ} = R (\frac{1}{2^{2}} - \frac{1}{4^{2}}) = \frac{3 R}{16}$ .....(2)

Divide equation (1) with (2)

$\frac{λ}{660} = \frac{5 \times 16}{36 \times 3}$

$λ = \frac{4400}{9} = 488 .88 = 488 .9 nm$