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Q. Assertion In the Lyman series of H-spectra, the maximum wavelength of lines is 121.65 nm. Reason Wavelength is maximum if there is transition from the very next level.

VMMC MedicalVMMC Medical 2009

Solution:

$ \frac{1}{\lambda }={{R}_{H}}{{Z}^{2}}\left[ \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right] $ For Lyman series, $ {{n}_{1}}=1 $ and $ {{n}_{2}}=2 $ [ $ \because $ for maximum $ \lambda , $ transition is from very next level.] $ \therefore $ $ \frac{1}{\lambda }=109678\times {{(1)}^{2}}\left[ \frac{1}{{{(1)}^{2}}}-\frac{1}{{{(2)}^{2}}} \right]c{{m}^{-1}} $ or $ =\frac{109678\times 3}{4}c{{m}^{-1}} $ $ \lambda =\frac{4\times {{10}^{7}}\times {{10}^{-7}}}{3\times 109678}\,cm=121.65\,nm. $