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Question

The first line of Balmer series has wavelength 6563 ˙A. What will be the wavelength of the first line of Lyman series?

A
1215.4 ˙A
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B
2500 ˙A
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C
7500 ˙A
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D
6563 ˙A
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Solution

The correct option is A 1215.4 ˙A

We know that,

1λ=R[1n211n22]

The first line of Balmer series corresponds to the transition from n2=3n1=2

1λB=R(122132)

1λB=R(1419)

1λB=5R36 .....(1)

Similarly,
The first line of Lyman series corresponds to the transition from n2=2n1=1

1λL=R(112122)

1λL=3R4 .....(2)

Using (1) and (2),

1/λB1/λL=(5R36)(3R4)=53×19

λLλB=527

λL=527λB=527×6563 ˙A

λL=1215.4 ˙A

Hence, option (A) is correct.

Why this question?
This question is very useful to understand the emission spectra of hydrogen atom, which very important concept for JEE Mains.

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