A certain species of ionised atoms produce emission line spectrum according to the Bohr model. A group of lines in the spectrum is forming a series in which the shortest wavelength is $ 22.79 \mathrm{~nm} $ and the longest wavelength is $ 41.02 \mathrm{~nm} $. The atomic number of atom is $ Z $ is

Question

A certain species of ionised atoms produce emission line spectrum according to the Bohr model. A group of lines in the spectrum is forming a series in which the shortest wavelength is $ 22.79 \mathrm{~nm} $ and the longest wavelength is $ 41.02 \mathrm{~nm} $. The atomic number of atom is $ Z $ is, $ 2 $, $ 3 $, $ 4 $, $ 5 $

MARKS App · Accepted Answer

Let n be the lowest energy level of the given series of lines, then maximum wavelength is given by h c λ m a x = 13.6 × Z 2 1 n 2 - 1 n + 1 2 h c λ m i n = 13.6 × Z 2 1 n 2 - 1 ∞ 2 Where λ m a x = 41.02 n m and λ m i n = 22.79 n m . Solving above equations, we get n = 2 , which corresponds to Balmer series. Z = 4

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