Identify the pair of species having the same | vedclass

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(B) For a single-electron species, the energy is given by the formula: $E_{n} = -13.6 	imes \frac{Z^2}{n^2} \ 	ext{eV}$.For $Li^{2+}$, the atomic nu

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$Li^{2+}(n=3), Be^{3+}(n=4)$

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(B) For a single-electron species, the energy is given by the formula: $E_{n} = -13.6 	imes \frac{Z^2}{n^2} \ 	ext{eV}$.For $Li^{2+}$, the atomic number $Z = 3$. At $n = 3$, the energy is $E_3 = -13.6 	imes \frac{3^2}{3^2} = -13.6 \ 	ext{eV}$.For $Be^{3+}$, the atomic number $Z = 4$. At $n = 4$, the energy is $E_4 = -13.6 	imes \frac{4^2}{4^2} = -13.6 \ 	ext{eV}$.Since both species have the same energy of $-13.6 \ 	ext{eV}$, the correct pair is $Li^{2+}(n=3)$ and $Be^{3+}(n=4)$.

Identify the pair of species having the same energy from the following (The number given in the bracket corresponds to the principal quantum number $(n)$ in which the electron is present).

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