The wavelength of the first member of Lyman series is 1216 Å. Calculate the wavelength of second member of Balmer series.

Solution

Given,
Wavelength of the first member of lyman series = 1216 Å

Now, the rydberg's formula gives us,

$\frac{1}{λ} = R (\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}})$

For first member of Lyman series, $n_{1} = 1 and n_{2} = 2 .$
$∴$    $\frac{1}{λ_{1}} = R (\frac{1}{1^{2}} - \frac{1}{4})$

$\Rightarrow$    $\frac{1}{λ_{1}} = \frac{3 R}{4}$

$\Rightarrow$    $λ_{1} = \frac{4}{3 R}$ ...(i)

For second member of Balmer series, $n_{1} = 2, n_{2} = 4$

Therefore,
   $\frac{1}{λ_{2}} = R (\frac{1}{2^{2}} - \frac{1}{4^{2}}) = \frac{3 R}{16}$

$\Rightarrow$    $λ_{2} = \frac{16}{3 R}$ ...(ii)

Dividing equation (ii) by (i), we get

$\frac{λ_{2}}{λ_{1}} = \frac{16}{3 R} \times \frac{3 R}{4} = 4$

$\Rightarrow$ $λ_{2} = 4 λ_{1} = 4 \times 1216 Å = 4864 Å .$

is the wavelength of the second member of balmer series.