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Question

The wavelength of the first spectral line in the Balmer series of hydrogen atom is 6561 ˙A. The wavelength of the second spectral line in the Balmer series of single ionized helium atom is ____ ˙A.



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Solution

The Rydberg's formula for the wavelength of the spectral lines is given by,

1λ=Z2R(1nf21ni2)

For H- atom, Z=1

The first line of Balmer series corresponds to the transition from ni=3nf=2

1λH=R(1)2(1419)

1λH=R(536) ..........(1)

For single ionized He- atom, Z=2

The second line of Balmer series corresponds to the transition from ni=4nf=2

1λHe=R(2)2(14116)

1λHe=R(4)(316) ........(2)

Dividing eq. (1) by (2),

λHeλH=14[163×536]=527

λHe=527×6561 ˙A=1215 ˙A

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