- Home
- Standard 12
- Physics
13.Nuclei
medium
Radius of $_2H{e^4}$ nucleus is $3 \,Fermi.$ The radius of $_{82}P{b^{206}}$ nucleus will be..........$fermi$
A
$5$
B
$6$
C
$11.16$
D
$8 $
Solution
(c) We have $r \propto {A^{1/3}} \Rightarrow \frac{{{r_2}}}{{{r_1}}} = {\left( {\frac{{{A_2}}}{{{A_1}}}} \right)^{1/3}} = {\left( {\frac{{206}}{4}} \right)^{1/3}}$
$\therefore {r_2} = 3{\left( {\frac{{206}}{4}} \right)^{1/3}} = 11.6\;Fermi$.
Standard 12
Physics
