The first member of the Balmer series of hydrogen atom has a wavelength

Question:

The first member of the Balmer series of hydrogen atom has a wavelength of $6561 \AA$. The wavelength of the second member of the Balmer series (in $\mathrm{nm}$ ) is_________

Solution:

The wavelength of the spectral line of hydrogen spectrum is given by formula

$\frac{1}{\lambda}=R\left(\frac{1}{n_{f}^{2}}-\frac{1}{n_{i}^{2}}\right)$

Where, $R=$ Rydberg constant

For the first member of Balmer series $n_{\mathrm{F}}=2, n_{\mathrm{i}}=3$

$\therefore \frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right)$            ...(1)

For last member of Balmer series, $n_{\mathrm{f}}=2, n_{\mathrm{i}}=4$

So, $\frac{1}{\lambda^{\prime}}=R\left[\frac{1}{4}-\frac{1}{16}\right]$                               ...(2)

Dividing (i) by (ii), we get

$\Rightarrow \quad \frac{\lambda^{\prime}}{\lambda}=\frac{5 \times 16}{9 \times 4 \times 3}$

$\Rightarrow \lambda^{\prime}=\frac{5 \times 4 \times 656.1}{9 \times 3}(\mathrm{~nm})=486 \mathrm{~nm}$

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