13.Nuclei
medium

If the nuclear radius of ${}_{13}^{27}Al$ is $3.6$ fermi the approximate nuclear radius of ${}_{}^{64}Cu$ in fermi is ..........$fm$

A

$2.4 $

B

$1.2 $

C

$4.8$

D

$3.6$

(AIPMT-2012)

Solution

Nuclear radius, $R=R_{0} A^{1 / 3}$

where $R_{0}$ is a constant and $A$ is the mass number

$\therefore \quad \frac{R_{\mathrm{Al}}}{R_{\mathrm{Cu}}}=\frac{(27)^{1 / 3}}{(64)^{1 / 3}}=\frac{3}{4}$

or $\quad R_{\mathrm{Cu}}=\frac{4}{3} \times R_{\mathrm{Al}}=\frac{4}{3} \times 3.6 \,fermi =4.8\,fermi$

Standard 12
Physics

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