Which of the following statement(s) is(are) c | vedclass

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(B) For $(A)$: For the Balmer series,transitions occur to $n=2$. The longest wavelength corresponds to $n=3 	o n=2$,and the shortest to $n=\infty 	o

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$A, C$

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(B) For $(A)$: For the Balmer series,transitions occur to $n=2$. The longest wavelength corresponds to $n=3 	o n=2$,and the shortest to $n=\infty 	o n=2$.$\frac{1}{\lambda_{\max}} = R(\frac{1}{2^2} - \frac{1}{3^2}) = R(\frac{5}{36}) \Rightarrow \lambda_{\max} = \frac{36}{5R}$.$\frac{1}{\lambda_{\min}} = R(\frac{1}{2^2} - 0) = \frac{R}{4} \Rightarrow \lambda_{\min} = \frac{4}{R}$.Ratio $\frac{\lambda_{\max}}{\lambda_{\min}} = \frac{36/5R}{4/R} = \frac{9}{5}$. Statement $(A)$ is correct.For $(B)$: Balmer range is $[364.6 \ nm, 656.3 \ nm]$. Paschen range is $[820.4 \ nm, 1875.1 \ nm]$. They do not overlap. Statement $(B)$ is incorrect.For $(C)$: For Lyman series,$\frac{1}{\lambda} = R(1 - \frac{1}{m^2})$ where $m=2, 3, \dots$. Since $\frac{1}{\lambda_0} = R$,we have $\frac{1}{\lambda} = \frac{1}{\lambda_0}(1 - \frac{1}{m^2}) \Rightarrow \lambda = \frac{\lambda_0}{1 - 1/m^2}$. Statement $(C)$ is correct.For $(D)$: Lyman range is $[91.2 \ nm, 121.6 \ nm]$. Balmer range is $[364.6 \ nm, 656.3 \ nm]$. They do not overlap. Statement $(D)$ is correct.

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