Radius of _2H{e^4} nucleus is 3 ,Fermi. The r | vedclass

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Question

(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}} \r

Vedclass · Accepted Answer

$11.16$

Vedclass · Answer

(c) We have $r \propto {A^{1/3}} \Rightarrow \frac{{{r_2}}}{{{r_1}}} = {\left( {\frac{{{A_2}}}{{{A_1}}}} ight)^{1/3}} = {\left( {\frac{{206}}{4}} ight)^{1/3}}$

$	herefore {r_2} = 3{\left( {\frac{{206}}{4}} ight)^{1/3}} = 11.6\;Fermi$.

Radius of $_2H{e^4}$ nucleus is $3 \,Fermi.$ The radius of $_{82}P{b^{206}}$ nucleus will be..........$fermi$

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