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(B) For a hydrogen atom,the energy of the ground state $(n=1)$ is $E_1 = -\frac{13.6}{1^2} = -13.6 	ext{ eV}$.For the second excited state,$n=3$. The

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$(a) \sim 12 	ext{ eV}, (b) 13.6 	ext{ eV}$

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(B) For a hydrogen atom,the energy of the ground state $(n=1)$ is $E_1 = -\frac{13.6}{1^2} = -13.6 	ext{ eV}$.For the second excited state,$n=3$. The energy is $E_3 = -\frac{13.6}{3^2} = -\frac{13.6}{9} \approx -1.51 	ext{ eV}$.The energy required to excite the electron from the ground state to the second excited state is $\Delta E = E_3 - E_1 = -1.51 - (-13.6) = 12.09 	ext{ eV} \approx 12 	ext{ eV}$.$(b)$ The ionization energy is the energy required to remove the electron from the ground state $(n=1)$ to infinity $(n=\infty)$.$E_{\infty} = 0 	ext{ eV}$.Energy required = $E_{\infty} - E_1 = 0 - (-13.6) = 13.6 	ext{ eV}$.

Energy of a stationary electron in the hydrogen atom is $E = -\frac{13.6}{n^2} \text{ eV}$. Calculate the energies required to excite the electron in a hydrogen atom to $(a)$ its second excited state and $(b)$ its ionized state,respectively.

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