The radius of the nucleus of { }_{13}^{27} Al | vedclass

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(C) The radius of a nucleus is given by the formula $R = R_{0} A^{1/3}$,where $R_{0} = 1.2 	ext{ fm} = 1.2 	imes 10^{-15} 	ext{ m}$ and $A$ is the

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$3.6 	imes 10^{-15} 	ext{ m}$

Vedclass · Answer

(C) The radius of a nucleus is given by the formula $R = R_{0} A^{1/3}$,where $R_{0} = 1.2 	ext{ fm} = 1.2 	imes 10^{-15} 	ext{ m}$ and $A$ is the mass number.For ${ }_{13}^{27} Al$,the mass number $A = 27$.Substituting the values into the formula:$R = 1.2 	imes 10^{-15} 	imes (27)^{1/3} 	ext{ m}$Since $(27)^{1/3} = 3$,we get:$R = 1.2 	imes 10^{-15} 	imes 3 	ext{ m}$$R = 3.6 	imes 10^{-15} 	ext{ m}$Thus,the correct option is $C$.

The radius of the nucleus of ${ }_{13}^{27} Al$ is $\qquad$ . (Given: $R_{0}=1.2 \text{ fm}$)

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