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JEE Main levelSingle correct optionDifficulty 2 of 3as072

Use $R_{H} = 1.097 \times 1 0^{7} m^{- 1}$ . The wavelength of the first line (longest wavelength) of the Lyman series of hydrogen is

A $91.2 nm$
B $121.5 nm$
C $656.3 nm$
D $102.6 nm$

Answer: B

1/ λ = R_{H} (1/ 1^{2} - 1/ 2^{2}) = 3 R_{H} /4

, so

λ = 4/ (3 R_{H}) = 121.5 nm

. The other values correspond to the series limit, H-alpha, and the

3 \to 1

line.

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