The total energy of the electron of H atom in | vedclass

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Question

(C) The energy of an electron in a hydrogen-like species is given by the formula: $E_n = -13.6 	imes \frac{Z^2}{n^2} \ eV$. For the $H$ atom $(Z=1)$

Vedclass · Accepted Answer

$-\frac{16}{9}E_2$

Vedclass · Answer

(C) The energy of an electron in a hydrogen-like species is given by the formula: $E_n = -13.6 	imes \frac{Z^2}{n^2} \ eV$. For the $H$ atom $(Z=1)$ in the second quantum state $(n=2)$: $E_2 = -13.6 	imes \frac{1^2}{2^2} = -\frac{13.6}{4} \ eV$. For the $He^{+}$ ion $(Z=2)$ in the third quantum state $(n=3)$: $E_3 = -13.6 	imes \frac{2^2}{3^2} = -13.6 	imes \frac{4}{9} \ eV$. Comparing the two expressions: $\frac{E_3}{E_2} = \frac{-13.6 	imes (4/9)}{-13.6 	imes (1/4)} = \frac{4}{9} 	imes 4 = \frac{16}{9}$. Wait,let us re-evaluate: $E_2 = -\frac{13.6}{4}$ and $E_3 = -\frac{13.6 	imes 4}{9}$. Thus,$E_3 = E_2 	imes \frac{4/9}{1/4} = E_2 	imes \frac{16}{9}$. Since the question defines the energy as $-E_2$,the magnitude is $E_2$. The correct value is $-\frac{16}{9}E_2$.

The total energy of the electron of $H$ atom in the second quantum state is $-E_2$. The total energy of the $He^{+}$ ion in the third quantum state is

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