JEE MAIN - Chemistry (2020 - 8th January Morning Slot - No. 6)
For the Balmer series in the spectrum of H atom,
$$\overline \nu = {R_H}\left\{ {{1 \over {n_1^2}} - {1 \over {n_2^2}}} \right\}$$, 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
$$\overline \nu = {R_H}\left\{ {{1 \over {n_1^2}} - {1 \over {n_2^2}}} \right\}$$, 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
(II), (III), (IV)
(I), (II), (III)
(I), (III), (IV)
(I), (II), (IV)
Explanation
For balmer series : n1 = 2, n2 = 3, 4, 5, .....$$\infty $$
For longest wavelength n2 = 3
$${1 \over \lambda } = R\left( {{1 \over {{2^2}}} - {1 \over {{3^2}}}} \right)$$
As wavelength decreases the lines in the Balmer series converge. The correct statements are (I), (II) and (III).
For longest wavelength n2 = 3
$${1 \over \lambda } = R\left( {{1 \over {{2^2}}} - {1 \over {{3^2}}}} \right)$$
As wavelength decreases the lines in the Balmer series converge. The correct statements are (I), (II) and (III).
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