In a series in the line spectrum of hydrogen, the wavelength of radiation is 6,563A^o. The name of the series and the orbits in which electron transition takes place are

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Paschen series, 4^th to 3^rd orbit

Pfund series, 6^th to 5^th orbit

Lyman series, 2^ndto 1^storbit

Balmer series, 3^rd to 2^nd orbit

answer is A.

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Detailed Solution

$\frac{1}{λ} = R Z^{2} [\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}] \frac{1}{6569 \times 10^{- 8}} = 109677.8 \times 1^{2} [\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}]$

On solving ,

n₁=2 and n₂ = 3

It is first line in balmer series.

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