Question

The radius of the oxygen nucleus ${}_8{O}^{16}$ is 2.8 x ${10}^{-15}m$. Then the radius of the lead nucleus ${}_{82}P{b}^{205}$ is

• A. 6.15 x ${10}^{-15}m$
• B. 6.55 x ${10}^{-15}m$
• C. 6.05 x ${10}^{-15}m$
• D. 6.75 x ${10}^{-15}m$

Answer 1

Answer: (B)

6.55 x ${10}^{-15}m$

Using Fermi model, the radius of a nucleus is related to the mass number, A as

$r\propto A^{1/3}$

Hence, $\frac{r_{Pb}}{r_O}=(\frac{205}{16}{)}^{1/3}=2.34$

Hence, $r_{Pb}=6.55\times {10}^{-15}m$