The ratio of the energy of the He^{+} ion in | vedclass

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(A) The energy of an electron in a hydrogen-like species is given by $E_n = -13.6 	imes \frac{Z^2}{n^2} 	ext{ eV}$.For $He^{+}$,the atomic number $Z

Vedclass · Accepted Answer

$9 : 4$

Vedclass · Answer

(A) The energy of an electron in a hydrogen-like species is given by $E_n = -13.6 	imes \frac{Z^2}{n^2} 	ext{ eV}$.For $He^{+}$,the atomic number $Z = 2$.The first excited state corresponds to $n = 2$,and the second excited state corresponds to $n = 3$.Energy in the first excited state $(E_1)$ = $-13.6 	imes \frac{2^2}{2^2} = -13.6 	ext{ eV}$.Energy in the second excited state $(E_2)$ = $-13.6 	imes \frac{2^2}{3^2} = -13.6 	imes \frac{4}{9} 	ext{ eV}$.The ratio of energy in the first excited state to the second excited state is $\frac{E_1}{E_2} = \frac{-13.6 	imes (4/4)}{-13.6 	imes (4/9)} = \frac{1}{4/9} = \frac{9}{4}$.Thus,the ratio is $9 : 4$.

The ratio of the energy of the $He^{+}$ ion in its first and second excited states will be:

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