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(C) The radius of the $n^{th}$ Bohr orbit is given by $r_n = 52.9 	imes \frac{n^2}{Z} \ pm$. For $He^{+}$, the atomic number $Z = 2$. Radius of the f

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$1.85 	imes 10^{-10}$

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(C) The radius of the $n^{th}$ Bohr orbit is given by $r_n = 52.9 	imes \frac{n^2}{Z} \ pm$. For $He^{+}$, the atomic number $Z = 2$. Radius of the fourth orbit $(n=4)$: $r_4 = 52.9 	imes \frac{4^2}{2} = 52.9 	imes 8 = 423.2 \ pm$. Radius of the third orbit $(n=3)$: $r_3 = 52.9 	imes \frac{3^2}{2} = 52.9 	imes 4.5 = 238.05 \ pm$. Difference in radii: $\Delta r = r_4 - r_3 = 423.2 - 238.05 = 185.15 \ pm$. Converting to meters: $185.15 	imes 10^{-12} \ m = 1.8515 	imes 10^{-10} \ m$.

The difference in radii between the fourth and third Bohr orbits of $He^{+}$ (in $m$) is:

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