Estimate the approximate volume of an alumini | vedclass

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(A) The radius of a nucleus is given by $R = R_0 A^{1/3}$.The volume of a nucleus is $V = \frac{4}{3} \pi R^3$.Substituting $R$,we get $V = \frac{4}{3

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$1 	imes 10^{-13} \ (	ext{Å})^3$

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(A) The radius of a nucleus is given by $R = R_0 A^{1/3}$.The volume of a nucleus is $V = \frac{4}{3} \pi R^3$.Substituting $R$,we get $V = \frac{4}{3} \pi (R_0 A^{1/3})^3 = \frac{4}{3} \pi R_0^3 A$.Given $R_0 \simeq 1.0 	imes 10^{-15} \ m$,$\pi \simeq 3$,and $A = 27$:$V = \frac{4}{3} 	imes 3 	imes (1.0 	imes 10^{-15} \ m)^3 	imes 27$.$V = 4 	imes 10^{-45} 	imes 27 \ m^3 = 108 	imes 10^{-45} \ m^3$.Since $1 \ 	ext{Å} = 10^{-10} \ m$,then $1 \ m = 10^{10} \ 	ext{Å}$,so $1 \ m^3 = 10^{30} \ (	ext{Å})^3$.$V = 108 	imes 10^{-45} 	imes 10^{30} \ (	ext{Å})^3 = 108 	imes 10^{-15} \ (	ext{Å})^3 \approx 1 	imes 10^{-13} \ (	ext{Å})^3$.

Estimate the approximate volume of an aluminium nucleus $(A=27)$.
$\text{Use } (R_0 \simeq 1.0 \times 10^{-15} \ m, \pi \simeq 3)$.

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Estimate the approximate volume of an aluminium nucleus $(A=27)$.$\text{Use } (R_0 \simeq 1.0 \times 10^{-15} \ m, \pi \simeq 3)$.