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(B) The energy of an electron in the $n^{th}$ shell of a hydrogen-like species is given by $E_n = -13.6 	imes \frac{Z^2}{n^2} 	ext{ eV}$.For $He^{+}

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$6.04$

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(B) The energy of an electron in the $n^{th}$ shell of a hydrogen-like species is given by $E_n = -13.6 	imes \frac{Z^2}{n^2} 	ext{ eV}$.For $He^{+}$,the atomic number $Z = 2$.The second excited state corresponds to $n = 3$ (since $n=1$ is ground state,$n=2$ is first excited state,and $n=3$ is second excited state).The total energy $E_3 = -13.6 	imes \frac{2^2}{3^2} = -13.6 	imes \frac{4}{9} \approx -6.04 	ext{ eV}$.Since the kinetic energy $(K.E.)$ is equal to the negative of the total energy $(K.E. = -E_n)$,we have $K.E. = -(-6.04 	ext{ eV}) = 6.04 	ext{ eV}$.

What is the kinetic energy of an electron in the second excited state of $He^{+}$? (Energy of an electron in the first shell of $H = -13.6 \text{ eV}$)

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