For the Balmer series in the spectrum of H atom, $\bar{v}=\mathrm{R}_{\mathrm{H}}\left\{\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}\right\}$, 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
Select the correct option:
A
(I), (II), (III)
B
(II), (III), (IV)
C
(I), (III), (IV)
D
(I), (II), (IV)
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
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.
n1 = 2 and n2 = 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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