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(C) The atomic density of $Si$ is $5 	imes 10^{28} 	ext{ atoms/m}^3$. Since $1 	ext{ m}^3 = 10^6 	ext{ cm}^3$,the density in $	ext{cm}^{-3}$ is $

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$1 	imes 10^{15} 	ext{ atoms/cm}^3$

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(C) The atomic density of $Si$ is $5 	imes 10^{28} 	ext{ atoms/m}^3$. Since $1 	ext{ m}^3 = 10^6 	ext{ cm}^3$,the density in $	ext{cm}^{-3}$ is $5 	imes 10^{28} / 10^6 = 5 	imes 10^{22} 	ext{ atoms/cm}^3$.Given that one indium atom is added per $5 	imes 10^7$ silicon atoms,the number of indium (acceptor) atoms per $	ext{cm}^3$ is calculated as:$n = \frac{5 	imes 10^{22}}{5 	imes 10^7} = 1 	imes 10^{15} 	ext{ atoms/cm}^3$.Thus,there are $1 	imes 10^{15}$ acceptor atoms in a $1 	ext{ cm}^3$ cube of silicon.

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