Calculate the wavenumber for the longest wavelength transition in the Balmer series of atomic hydrogen.

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Calculate the wavenumber for the longest wavelength transition in the Balmer series of atomic hydrogen.

The Balmer series of the hydrogen emission spectrum, n_i = 2.

Hence, wavenumber expression $ν$ is:

$\bar{\nu}=\left[\frac{1}{(2)^{2}}-\frac{1}{n_{f}^{2}}\right]\left(1.097 \times 10^{7} \mathrm{~m}^{-1}\right)$
Since wave number $(\bar{\nu})$ is inversely proportional to the transition wavelength, the lowest possivle value of $(\bar{\nu})$ corresponds to the longest wavelength transition.

For $(\bar{\nu})$ to be of the lowest possible value, $n_{f}$ should be minimum. In the Balmer series, transitions from $n_{i}=2$ to $\mathrm{n}_{\mathrm{f}}=3$ are allowed.
Hence, taking $\mathrm{n}_{\mathrm{f}}=3$ , we get:
$\begin{array}{l} \bar{\nu}=\left(1.097 \times 10^{7}\right)\left[\frac{1}{(2)^{2}}-\frac{1}{3^{2}}\right] \\ \bar{\nu}=\left(1.097 \times 10^{7}\right)\left[\frac{1}{4}-\frac{1}{9}\right] \\ =\left(1.097 \times 10^{7}\right)\left[\frac{9-4}{36}\right] \\ =\left(1.097 \times 10^{7}\right)\left[\frac{5}{36}\right] \\ \bar{\nu}=1.5236 \times 10^{6} \mathrm{~m}^{-1} \end{array}$