For the Balmer series in the spectrum of H atom, $\bar{\nu} = R_H\left\{\frac{1}{n_1^2}-\frac{1}{n_2^2}\right\}$, 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 wave

Question

For the Balmer series in the spectrum of H atom, $\bar{\nu} = R_H\left\{\frac{1}{n_1^2}-\frac{1}{n_2^2}\right\}$, 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 (1) (II), (III), (IV) (2) (I), (III), (IV) (3) (I), (II), (III) (4) (I), (II), (IV)

Accepted Answer

(2) (I), (III), (IV). The correct option includes statements that accurately describe the Balmer series: as the wavelength decreases, the spectral lines converge due to higher energy transitions, the longest wavelength corresponds to the transition from n=3 to n=2, and the ionization energy can indeed be derived from the wave numbers of these transitions.

Solution

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