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(C) The radius of the $n$-th orbit in a hydrogen atom is given by $r_n = a_0 n^2$,where $a_0$ is the Bohr radius.Therefore,the ratio of the radii is $

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$486$

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(C) The radius of the $n$-th orbit in a hydrogen atom is given by $r_n = a_0 n^2$,where $a_0$ is the Bohr radius.Therefore,the ratio of the radii is $r_i / r_f = n_i^2 / n_f^2 = 16 / 4 = 4$.Taking the square root,we get $n_i / n_f = 2$,which implies $n_i = 2n_f$.For the simplest transition,we take $n_f = 2$ and $n_i = 4$.Using the Rydberg formula for the wavelength $\lambda$:$1/\lambda = R(1/n_f^2 - 1/n_i^2)$$1/\lambda = R(1/2^2 - 1/4^2) = R(1/4 - 1/16) = R(3/16)$.Substituting $R = 1.0973 	imes 10^7 	ext{ m}^{-1}$:$1/\lambda = 1.0973 	imes 10^7 	imes (3/16) \approx 0.20574 	imes 10^7 	ext{ m}^{-1}$.$\lambda = 1 / (0.20574 	imes 10^7) \approx 4.86 	imes 10^{-7} 	ext{ m} = 486 	ext{ nm}$.

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