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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