For the Balmer series in the spectrum of H atom, υ¯$$=RH1n12$$−$$1n22$$, the correct statements among I to IV areI. As wavelength decreases, the lines in the series converge.II. The integer $$n1$$ is equal to $$2$$.III. The lines of longest wavelength corresponds to $$n2$$ $$=$$ $$3$$.IV. The ionisation energy of hydrogen can be calculated from wave number of these lines.

Question

For the Balmer series in the spectrum of H atom, υ¯$$=RH1n12$$−$$1n22$$, the correct statements among I to IV areI. As wavelength decreases, the lines in the series converge.II. The integer $$n1$$ is equal to $$2$$.III. The lines of longest wavelength corresponds to $$n2$$ $$=$$ $$3$$.IV. The ionisation energy of hydrogen can be calculated from wave number of these lines.

Infinity Learn · Accepted Answer

Statements I, II and III are correct. The explanations of these statements are as followsI. In each series of hydrogen spectrum, lines converge at the higher frequency or lower wavelength limit.∴This is correct.II. For Balmer series, $$n1$$ $$=$$ $$2$$∴This is correct.III. Longest wavelength represents the smallest wave numberv¯$$=RH1n12$$−$$1n22$$⋅$$n2$$>$$n1When$$ $$n1$$ $$=$$ $$2$$, smallest value of v¯  is for $$n2$$ $$=$$ $$3$$∴This is a correct statement.IV. This statement is not correct. It's corrected from is as follows : Ionisation energy of hydrogen can be calculated from Lyman series lines (n$$ $$=$$ $$1) only by using the following formula, v¯$$=RH1(1)2$$−$$1$$∞$$2$$.

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