What is the kinetic energy of an electron in | vedclass

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

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$6.04$

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(B) For a hydrogen-like species,the energy of an electron in the $n^{th}$ shell is given by $E_n = -13.6 	imes \frac{Z^2}{n^2} \ eV$. For $He^{+}$,the atomic number $Z = 2$. The second excited state corresponds to $n = 3$ (ground state $n=1$,first excited state $n=2$,second excited state $n=3$). Substituting the values: $E_3 = -13.6 	imes \frac{2^2}{3^2} = -13.6 	imes \frac{4}{9} \ eV$. $E_3 = -6.044 \ eV$. The kinetic energy $(KE)$ of an electron is equal to the magnitude of its total energy,i.e.,$KE = |E_n|$. Therefore,$KE = 6.044 \ eV \approx 6.04 \ eV$.

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

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