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Using the Rydberg formula. calculate the wavelengths of the first four spectral lines in the Lyman series of the hydrogen spectrum.
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## QuestionPhysicsClass 12

Using the Rydberg formula. calculate the wavelengths of the first four spectral lines in the Lyman series of the hydrogen spectrum.

The Rydberg formula is
$$\mathit{hc}\text /\lambda _{\mathit{ij}}=\dfrac{me^4}{8\varepsilon _o^2h^2}\left(\dfrac 1{n_f^2}-\dfrac 1{n_i^2}\right)$$
The wavelengths of the first four lines in the Lyman series correspond to transitions from $$n_i=2.3.4.5$$  to $$n_f=1.$$  We know that $$\dfrac{me^4}{8\varepsilon _0^2h^2}=13.6\;\mathit{eV}=21.76\times 10^{-19}J$$
Therefore.
$$\lambda _{11}=\dfrac{\mathit{hc}}{21.76\times 10^{-19}\left(\dfrac 1 1-\dfrac 1{n_1^2}\right)}m$$
$$=\dfrac{6.625\times 10^{-34}\times 3\times 10^8\times n_1^2}{21.76\times 10^{-19}\times (n_i^2-1)}m=\dfrac{0.9134n_i^2}{(n_i^2-1)}\times 10^{-7}m$$
$$=913.4\;n_t^2/(n_t^2-1)\;Å$$
Substituting $$n_4=2.3,4.5.$$  we get
$$\lambda _{21}=1218\;Å$$   $$.\lambda _{31}=1028\text{A.}\lambda _{41}=974.3\;A$$  and $$\lambda _{51}=951.4\;Å$$