    Question

# The wavelength of the second line in Balmer series of the hydrogen spectrum is 486.4 nm. What is the wavelength of the first line of the Lyman series?

A
121.6 nm
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B
364.8 nm
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C
729.6 cm
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D
None of these
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Solution

## The correct option is A 121.6 nmFor Balmer series, 1λ=R(122−1n2)=R(14−1n2);n=3,4,5⋯ The second line in balmer series corresponds ot n= 4. Hence, 1λ2=R(14−116)=3R16⇒λ2=163R The wavelength of the first line (n = 2) in Lyman series is given by 1λ1=R(1−122)=R(1−14)=3R4⇒λ1=43R ∴λ1λ2=43R×3R16=14⇒λ1=λ24=486.44=121.6 nm  Suggest Corrections  0      Similar questions
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