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(D) For the hydrogen atom,the energy of a transition is given by $\Delta E = 13.6 \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} ight) 	ext{ eV}$.Case $1

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$27$

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(D) For the hydrogen atom,the energy of a transition is given by $\Delta E = 13.6 \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} ight) 	ext{ eV}$.Case $1$: Second excited state $(n_i = 3)$ to first excited state $(n_f = 2)$.$\frac{hc}{\lambda_0} = 13.6 \left( \frac{1}{2^2} - \frac{1}{3^2} ight) = 13.6 \left( \frac{1}{4} - \frac{1}{9} ight) = 13.6 \left( \frac{5}{36} ight) \ldots (i)$Case $2$: Third excited state $(n_i = 4)$ to second orbit $(n_f = 2)$.$\frac{hc}{\lambda'} = 13.6 \left( \frac{1}{2^2} - \frac{1}{4^2} ight) = 13.6 \left( \frac{1}{4} - \frac{1}{16} ight) = 13.6 \left( \frac{3}{16} ight) \ldots (ii)$Given $\lambda' = \frac{20}{x} \lambda_0$,we divide equation $(i)$ by $(ii)$:$\frac{\lambda'}{\lambda_0} = \frac{13.6 \left( \frac{5}{36} ight)}{13.6 \left( \frac{3}{16} ight)} = \frac{5}{36} 	imes \frac{16}{3} = \frac{5 	imes 4}{9 	imes 3} = \frac{20}{27}$Comparing $\frac{\lambda'}{\lambda_0} = \frac{20}{27}$ with $\frac{20}{x}$,we get $x = 27$.

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