The energy of an electron in the ground state | vedclass

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(D) The energy of an electron in a hydrogen-like species is given by the formula: $E_n = -R_H \left( \frac{Z^2}{n^2} \right) \ J$. For $He^{+}$ ion,$Z

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$-x$

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(D) The energy of an electron in a hydrogen-like species is given by the formula: $E_n = -R_H \left( \frac{Z^2}{n^2} \right) \ J$. For $He^{+}$ ion,$Z=2$ and $n=1$: $E_1 = -R_H \left( \frac{2^2}{1^2} \right) = -4 R_H$. Given $E_1 = -x \ J$,we have $-4 R_H = -x$,which implies $R_H = \frac{x}{4}$. For $Be^{3+}$ ion,$Z=4$ and $n=2$: $E_2 = -R_H \left( \frac{4^2}{2^2} \right) = -R_H \left( \frac{16}{4} \right) = -4 R_H$. Substituting $R_H = \frac{x}{4}$ into the expression for $E_2$: $E_2 = -4 \left( \frac{x}{4} \right) = -x \ J$.

The energy of an electron in the ground state $(n=1)$ for $He^{+}$ ion is $-x \ J$. What is the energy for an electron in the $n=2$ state for $Be^{3+}$ ion in $J$?

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