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(C) The wavelength of the spectral lines is given by $\lambda = \frac{1500p^2}{p^2 - 1}$.To find the ionization energy,we consider the transition from

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$8.28$

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(C) The wavelength of the spectral lines is given by $\lambda = \frac{1500p^2}{p^2 - 1}$.To find the ionization energy,we consider the transition from $n = \infty$ to $n = 1$,which corresponds to the series limit (minimum wavelength).As $p 	o \infty$,$\lambda_{\min} = \lim_{p 	o \infty} \frac{1500p^2}{p^2 - 1} = 1500 \mathring{A}$.The energy of the photon corresponding to this wavelength is $\Delta E = \frac{hc}{\lambda_{\min}}$.Substituting the given values,$\Delta E = \frac{12420 	ext{ eV-} \mathring{A}}{1500 \mathring{A}} = 8.28 	ext{ eV}$.Since the ionization potential is the energy required to remove an electron from the ground state $(n=1)$ to infinity,the ionization potential is $8.28 	ext{ V}$.

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