In the hydrogen atom spectrum, (R is Rydberg' | vedclass

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(B) The wavelength $\lambda$ for a transition is given by $\frac{1}{\lambda} = R \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)$.For the Lyman serie

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$A, B$ and $C$ Only

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(B) The wavelength $\lambda$ for a transition is given by $\frac{1}{\lambda} = R \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)$.For the Lyman series $(n_f = 1)$, the maximum wavelength occurs at $n_i = 2$: $\frac{1}{\lambda_{max}} = R \left( 1 - \frac{1}{4} \right) = \frac{3R}{4} \implies \lambda_{max} = \frac{4}{3R}$. Thus, statement $A$ is correct.The Balmer series $(n_f = 2)$ corresponds to the visible region. Thus, statement $B$ is correct.For the Paschen series $(n_f = 3)$, the minimum wavelength occurs at $n_i = \infty$: $\frac{1}{\lambda_{min}} = R \left( \frac{1}{3^2} - 0 \right) = \frac{R}{9} \implies \lambda_{min} = \frac{9}{R}$. Thus, statement $C$ is correct.For the Lyman series $(n_f = 1)$, the minimum wavelength occurs at $n_i = \infty$: $\frac{1}{\lambda_{min}} = R \left( 1 - 0 \right) = R \implies \lambda_{min} = \frac{1}{R}$. Thus, statement $D$ is incorrect.Therefore, statements $A, B,$ and $C$ are correct.

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