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.
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a
I, II, IV
b
II, III, IV
c
I, III, IV
d
I, II, III
answer is D.
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Detailed Solution
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.