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Energy of an electron in a particular orbit of single electron species of beryllium is the same as the energy of an electron in the ground state of hydrogen atom . Identify the orbit of beryllium

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Solution :Energy of an ELECTRON in `Be^(+3)` ion = `(-13.6)/(n^2)xx4^2ev`
Energy of the electron in ground state hydrogen atom = -13.6 EV
`THEREFORE(-13.6)/(n^2)xx4^2=-13.6`
`rArrn^2=42`
`n=42`


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