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(C) The energy of an electron in the $n^{th}$ orbit of a hydrogen-like species is given by $E_n = -13.6 	imes \frac{Z^2}{n^2} \ eV$. The energy diffe

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$5$

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(C) The energy of an electron in the $n^{th}$ orbit of a hydrogen-like species is given by $E_n = -13.6 	imes \frac{Z^2}{n^2} \ eV$. The energy difference between the $3^{rd}$ and $2^{nd}$ orbit is $\Delta E = E_3 - E_2 = -13.6 	imes Z^2 	imes (\frac{1}{3^2} - \frac{1}{2^2})$. $\Delta E = -13.6 	imes Z^2 	imes (\frac{1}{9} - \frac{1}{4}) = -13.6 	imes Z^2 	imes (\frac{4-9}{36}) = -13.6 	imes Z^2 	imes (-\frac{5}{36})$. $\Delta E = 13.6 	imes Z^2 	imes \frac{5}{36} = 1.888 	imes Z^2 \approx 1.89 	imes Z^2$. Given $\Delta E = 47.2 \ eV$,we have $47.2 = 1.89 	imes Z^2$. $Z^2 = \frac{47.2}{1.89} \approx 25$. Therefore,$Z = 5$.

$A$ single-electron species has a nuclear charge of $+Ze$,where $Z$ is a constant. The energy required to excite an electron from the $2^{nd}$ Bohr orbit to the $3^{rd}$ Bohr orbit is $47.2 \ eV$. Calculate the value of $Z$.

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