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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.As $p 	o \infty$,the wavelength $\lambda_{\min} = \lim_{p 	o \infty} \frac{1500p^2}{p^2 - 1} = 1500\ \mathring{A}$.The energy corresponding to this wavelength is the ionization energy $E_i$,given by $E_i = \frac{hc}{\lambda_{\min}}$.Substituting the given values,$E_i = \frac{12420\ eV\cdot\mathring{A}}{1500\ \mathring{A}} = 8.28\ eV$.Since the ionization energy is $8.28\ eV$,the ionization potential is $8.28\ V$.

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