Question
Given below are two statements:
Statement I: When an electric discharge is passed through gaseous hydrogen, the hydrogen molecules dissociate and the energetically excited hydrogen atoms produce electromagnetic radiation of discrete frequencies.
Statement II: The frequency of second line of Balmer series obtained from ${He}^{+}$is equal to that of first line of Lyman series obtained from hydrogen atom.
In the light of the above statements, choose the correct answer from the options given below:
Options
Solution
$\frac{1}{\lambda} = RZ^{2}\left( \frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}} \right)$
For Ist line of lyman series in H -atom
$${\frac{1}{\lambda} = R(1)^{2}\left( \frac{1}{1^{2}} - \frac{1}{2^{2}} \right) }{\frac{1}{\lambda} = \frac{3R}{4} }$$for $2^{\text{nd~}}$ line of Balmer series of ${He}^{+}$
$${\frac{1}{\lambda'} = R(2)^{2}\left( \frac{1}{2^{2}} - \frac{1}{4^{2}} \right) }{\frac{1}{\lambda'} = \frac{3R}{4} }$$As $\lambda$ and $\lambda'$ is equal so frequency of these lines will be also equal.
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