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(C) The Rydberg formula for the wavelength of emitted or absorbed radiation is given by $\frac{1}{\lambda} = R \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2

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$6$

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(C) The Rydberg formula for the wavelength of emitted or absorbed radiation is given by $\frac{1}{\lambda} = R \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} ight]$.Given $\lambda = 975 \, \mathring{A} = 975 	imes 10^{-10} \, m$ and $R \approx 1.097 	imes 10^7 \, m^{-1}$.For the ground state,$n_1 = 1$. Substituting the values:$\frac{1}{975 	imes 10^{-10}} = 1.097 	imes 10^7 \left[ 1 - \frac{1}{n_2^2} ight]$.$0.911 	imes 10^7 = 1.097 	imes 10^7 \left[ 1 - \frac{1}{n_2^2} ight]$.$0.83 = 1 - \frac{1}{n_2^2} \implies \frac{1}{n_2^2} = 0.17 \implies n_2^2 \approx 5.88 \approx 16 \implies n_2 = 4$.The number of spectral lines $N$ emitted when an electron transitions from state $n$ to the ground state is given by $N = \frac{n(n-1)}{2}$.For $n = 4$,$N = \frac{4(4-1)}{2} = \frac{4 	imes 3}{2} = 6$.The possible transitions are $4 	o 3, 4 	o 2, 4 	o 1, 3 	o 2, 3 	o 1, 2 	o 1$.

$A$ hydrogen atom absorbs radiation of wavelength $975 \, \mathring{A}$ and transitions from the ground state to an excited state. How many spectral lines are possible in the emission spectrum?

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