For the Balmer series in the spectrum of H atom, , the correct statements among (I) to (IV) are :(I) 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 ionization energy of hydrogen can be calculated from wave number of these lines

Engineering

Chemistry

Hydrogen Spectrum

Question

For the Balmer series in the spectrum of H atom, $\bar{v} = R_{H} [\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}]$ , the correct statements among (I) to (IV) are :

(I) 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 ionization energy of hydrogen can be calculated from wave number of these lines

(I), (III), (IV)

(I), (II), (IV)

(II), (III), (IV)

(I), (II), (III)

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The Balmer series describes visible spectral lines of hydrogen when electrons transition to n=2 from higher levels (n≥3). The wave number formula is: $\bar{v} = R_{H} [\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}]$ .

(I) True: As n₂ increases, energy differences decrease, so lines converge at higher frequency (shorter wavelength).

(II) True: For Balmer series, n₁ is always 2.

(III) True: Longest wavelength (lowest energy) comes from smallest transition: n₂=3 to n₁=2.

(IV) False: Ionization energy requires n₂→∞ transition, which is not part of this specific series formula.

Final answer: (I), (II), (III)