Use R_H = 1.097x10^7 m^-1. The Lyman series limit of a hydrogen atom lies at a longer wavelength than that of Li^2+. Find the difference between the two limits in nm (one decimal).

JEE Advanced levelNumerical answerDifficulty 3 of 3as101

Use $R_{H} = 1.097 \times 1 0^{7} m^{- 1}$ . The Lyman series limit of a hydrogen atom lies at a longer wavelength than that of $Li X^{2 +}$ . Find the difference between the two limits in nm (one decimal).

Numerical answer type, answer in nm, tolerance 0.1.

Series limit:

1/ λ = Z^{2} R_{H}

. H:

λ = 1/ R_{H} = 91.16 nm

Li X^{2 +}

λ = 1/ (9 R_{H}) = 10.13 nm

. Difference

= 81.0 nm