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Question

The wavelength of the first member of Balmer series is 6563 ∘A. Calculate the wavelength of second member of Lyman series.

A
1025.5 A
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B
2025 A
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C
6563 A
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D
None of these
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Solution

The correct option is A 1025.5 A
The first orbital of Balmer series corresponds to the transition from 3 to 2 and the second member of Lyman series corresponds to the transition from 3 to 1.

Let us write the expression for the wavelength for the first member of the Balmer series.

1λ1=RZ2(1n211n22)=RZ2(122132)
=RZ2(1419)=536RZ2 -------(1)

Let us write similar expression for the second member of Lyman series.

1λ2=RZ2(1n211n22)=RZ2(112132)
=RZ2(119)=89RZ2 -------(2)

Given that λ1=6563 A

Hence dividing equation (1) by (2),
λ26563=53689
And this gives λ2=1025.5 A

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