What is the separation energy (in eV) for Be^ | vedclass

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(D) For $Be^{3+}$,the atomic number $Z = 4$.The first excited state corresponds to $n = 2$.The energy of an electron in the $n^{th}$ orbit is given by

Vedclass · Accepted Answer

$54.4$

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(D) For $Be^{3+}$,the atomic number $Z = 4$.The first excited state corresponds to $n = 2$.The energy of an electron in the $n^{th}$ orbit is given by $E_n = -13.6 	imes \frac{Z^2}{n^2} \ eV$.For $n = 2$,$E_2 = -13.6 	imes \frac{4^2}{2^2} = -13.6 	imes \frac{16}{4} = -13.6 	imes 4 = -54.4 \ eV$.The separation energy (or ionization energy from the excited state) is the energy required to move the electron from $n = 2$ to $n = \infty$.$E_{separation} = E_{\infty} - E_2 = 0 - (-54.4 \ eV) = 54.4 \ eV$.

What is the separation energy (in $eV$) for $Be^{3+}$ in the first excited state?

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