The ground and first excited state energies a | vedclass

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

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$He^{+}(E_1), Be^{3+}(E_2)$

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(B) The energy of an electron in a hydrogen-like species is given by the formula $E_n = -13.6 	imes \frac{Z^2}{n^2} 	ext{ eV}$.For the ground state $(n=1)$,$E_1 = -13.6 	imes Z^2 	ext{ eV}$.For the first excited state $(n=2)$,$E_2 = -13.6 	imes \frac{Z^2}{4} 	ext{ eV}$.We need to check which pair has the same energy value.For option $D$: $H$ $(Z=1)$ in first excited state $(n=2)$ has $E_2 = -13.6 	imes \frac{1^2}{2^2} = -3.4 	ext{ eV}$.$Be^{3+}$ $(Z=4)$ in ground state $(n=1)$ has $E_1 = -13.6 	imes \frac{4^2}{1^2} = -13.6 	imes 16 = -217.6 	ext{ eV}$.Wait,let us re-evaluate the condition $E_n(Z_1) = E_m(Z_2)$.For $H(E_2)$,$n=2, Z=1$,$E = -13.6 	imes \frac{1}{4} = -3.4 	ext{ eV}$.For $Be^{3+}(E_1)$,$n=1, Z=4$,$E = -13.6 	imes \frac{16}{1} = -217.6 	ext{ eV}$.Actually,checking $He^{+}(E_1)$ $(Z=2, n=1)$ gives $E = -13.6 	imes 4 = -54.4 	ext{ eV}$.Checking $Be^{3+}(E_2)$ $(Z=4, n=2)$ gives $E = -13.6 	imes \frac{16}{4} = -54.4 	ext{ eV}$.Thus,$He^{+}(E_1)$ and $Be^{3+}(E_2)$ have the same energy.

The ground and first excited state energies are $E_1$ and $E_2$ respectively. Which pair of species has the same energy? [Note that energy is indicated in the bracket].

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