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^+$He^{+}$ 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:

### Related Formula frac1lambda = RZ^2left(frac1n_1^2 - frac1n_2^2right)$$\frac{1}{\lambda} = RZ^2\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$$ ### Core Logic Statement I is a factual description of how the hydrogen emission spectrum is obtained. Dissociation yields excited atoms that emit light at discrete frequencies. So Statement I is true. For Statement II: First line of Lyman series for H atom (Z=1, n_1=1, n_2=2$Z=1, n_1=1, n_2=2$): frac1lambda_1 = R(1)^2left(frac11^2 - frac12^2right) = Rleft(1 - frac14right) = frac3R4$$\frac{1}{\lambda_1} = R(1)^2\left(\frac{1}{1^2} - \frac{1}{2^2}\right) = R\left(1 - \frac{1}{4}\right) = \frac{3R}{4}$$ Second line of Balmer series for He^+$He^+$ (Z=2, n_1=2, n_2=4$Z=2, n_1=2, n_2=4$): frac1lambda_2 = R(2)^2left(frac12^2 - frac14^2right) = 4Rleft(frac14 - frac116right) = 4Rleft(frac316right) = frac3R4$$\frac{1}{\lambda_2} = R(2)^2\left(\frac{1}{2^2} - \frac{1}{4^2}\right) = 4R\left(\frac{1}{4} - \frac{1}{16}\right) = 4R\left(\frac{3}{16}\right) = \frac{3R}{4}$$ Since frac1lambda_1 = frac1lambda_2$\frac{1}{\lambda_1} = \frac{1}{\lambda_2}$, their wavelengths (and therefore frequencies) are exactly the same. Statement II is true. ### Step 1: Final Conclusion Both statements are correct. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Structure of Atom