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(D) For a hydrogen-like atom,the radius of the $n^{th}$ orbit is given by the formula:$r_n = \frac{n^2}{Z} r_0$where $r_0 = 0.51 	imes 10^{-10} 	ext

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$Be^{3+}$

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(D) For a hydrogen-like atom,the radius of the $n^{th}$ orbit is given by the formula:$r_n = \frac{n^2}{Z} r_0$where $r_0 = 0.51 	imes 10^{-10} 	ext{ m}$ is the radius of the first orbit of the hydrogen atom.Given the radius of the smallest orbit $(n=1)$ is $r_1 = \frac{0.51 	imes 10^{-10}}{4} 	ext{ m}$.Substituting these values into the formula:$\frac{0.51 	imes 10^{-10}}{4} = \frac{1^2}{Z} 	imes 0.51 	imes 10^{-10}$$\frac{1}{4} = \frac{1}{Z}$$Z = 4$The atomic number $Z=4$ corresponds to Beryllium $(Be)$. Since it is a hydrogen-like ion,it must be $Be^{3+}$.

The radius of the smallest electron orbit in a hydrogen-like ion is $(0.51/4 \times 10^{-10}) \text{ m}$. This ion is:

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