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(C) The wavelength of a spectral line in the hydrogen atom is given by the Rydberg formula: $\frac{1}{\lambda} = R \left( \frac{1}{n_f^2} - \frac{1}{n

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$4$

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(C) The wavelength of a spectral line in the hydrogen atom is given by the Rydberg formula: $\frac{1}{\lambda} = R \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)$.For the Balmer series,the final state is $n_f = 2$.Substituting the values,we have: $\frac{1}{\lambda} = R \left( \frac{1}{2^2} - \frac{1}{n_i^2} \right) = R \left( \frac{1}{4} - \frac{1}{n_i^2} \right)$.Given that $\lambda = \frac{16}{3 R}$,we substitute this into the equation:$\frac{1}{(16 / 3 R)} = R \left( \frac{1}{4} - \frac{1}{n_i^2} \right) \Rightarrow \frac{3 R}{16} = R \left( \frac{1}{4} - \frac{1}{n_i^2} \right)$.Dividing both sides by $R$,we get: $\frac{3}{16} = \frac{1}{4} - \frac{1}{n_i^2}$.Rearranging the terms to solve for $n_i^2$: $\frac{1}{n_i^2} = \frac{1}{4} - \frac{3}{16} = \frac{4 - 3}{16} = \frac{1}{16}$.Therefore,$n_i^2 = 16$,which gives $n_i = 4$.

The wavelength of a spectral line emitted by a hydrogen atom in the Balmer series is $\frac{16}{3 R}$ ($R$ is the Rydberg constant). What is the value of the principal quantum number of the state from which the transition takes place?

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