The series limit for Balmer series of H-spectra is

A
3800
B
4200
C
3646
D
4000

Solution:

We know that, For Balmer series,
$\frac{1}{λ} = R (\frac{1}{2^{2}} - \frac{1}{n_{2}^{2}})$
Where, $n_{2} = 3, 4, 5.... \infty$
For the limit of the series, $n_{2} = \infty$
The wavelength can be written as,
$\frac{1}{λ} = R (\frac{1}{2^{2}} - \frac{1}{\infty^{2}})$
Since $\frac{1}{\infty^{2}} \approx 0$ , we can write
$\frac{1}{λ} = R \frac{1}{2^{2}}$
From this we can write,
$R = \frac{4}{λ}$
From the question, we know
λ=3646A∘
The last line in any series is called series limit. Series limit for Balmer series is 3646 $\overset{o}{A}$ .

Solution:

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