| For the Balmer series in the spectrum of H atom,\[\overline{V}={{R}_{^{H}}}\left\{ \frac{1}{{{n}^{2}}_{1}}-\frac{1}{{{n}^{2}}_{2}} \right\},\] the correct statements among (I) to (VI) are |
| (I) as wavelength decreases, the lines in the series converge |
| (II) The integer \[{{n}_{1}}\] is equal to 2 |
| (III)The lines of longest wavelength corresponds to \[{{n}_{2}}=3\] |
| (IV)The ionization energy of hydrogen can be calculated from wave number of these lines |
A) (I), (II), (III)
B) (II), (III), (IV)
C) (I), (III), (IV)
D) (I), (II), (IV)
Correct Answer: A
Solution :
[a] In Balmer series of H-atom, the electronic transitions take place from higher orbits to 2nd orbit and the longest wavelength will correspond to transition from 3rd orbit to 2nd orbit. \[\therefore {{n}_{1}}=2\text{ }and\text{ }{{n}_{2}}=3\]For longest wavelength. As wavelength decreases the lines in the Balmer series converge. The correct statements are (I), (II) and (III).
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