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

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$5$

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(C) The energy of an electron in the $n^{th}$ orbit of a hydrogen-like atom is given by the formula:$E_n = -13.6 \frac{Z^2}{n^2} \ eV$where $Z$ is the atomic number and $n$ is the principal quantum number.The energy of the $2^{nd}$ orbit $(n=2)$ is:$E_2 = -13.6 \frac{Z^2}{2^2} = -13.6 \frac{Z^2}{4} \ eV$The energy of the $3^{rd}$ orbit $(n=3)$ is:$E_3 = -13.6 \frac{Z^2}{3^2} = -13.6 \frac{Z^2}{9} \ eV$The energy required to excite the electron from the $2^{nd}$ to the $3^{rd}$ orbit is given by $\Delta E = E_3 - E_2 = 47.2 \ eV$.Substituting the values:$47.2 = -13.6 \frac{Z^2}{9} - (-13.6 \frac{Z^2}{4})$$47.2 = 13.6 Z^2 \left( \frac{1}{4} - \frac{1}{9} ight)$$47.2 = 13.6 Z^2 \left( \frac{9-4}{36} ight)$$47.2 = 13.6 Z^2 \left( \frac{5}{36} ight)$$Z^2 = \frac{47.2 	imes 36}{13.6 	imes 5}$$Z^2 = \frac{1699.2}{68} \approx 24.988$$Z^2 \approx 25$$Z = 5$Therefore,the atomic number of the atom is $5$.

$A$ hydrogen-like atom has one electron revolving around a stationary nucleus. If the energy required to excite the electron from the $2^{nd}$ orbit to the $3^{rd}$ orbit is $47.2 \ eV$,find the atomic number of the given atom.

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