Question

The radius of ${}_{29}{Cu}^{64}$ nucleus in Fermi is (given $R_0=1.2\times {10}^{-15}m$)

• A. $4.8$
• B. $1.2$
• C. $7.7$
• D. $9.6$

Answer 1

Answer: (A)

$4.8$

Mass number of Cu nucleus $A=64$

Radius of ${}_{29}C{u}^{64}$ nucleus $R=R_o{A}^{1/3}$

$\therefore$ $R=1.2\times {10}^{-15}\times (64{)}^{1/3}=4.8\times {10}^{-15}$ $m$

$⟹$ $R=4.8$ $fm$