Gujarati
13.Nuclei
medium

$_2H{e^4}$ नाभिक की त्रिज्या $3$ फर्मी हैं, तो $_{82}P{b^{206}}$ नाभिक की त्रिज्या ..........$fermi$ होगी

A

$5$

B

$6 $

C

$11.16 $

D

$8$

Solution

हम जानते हैं $r \propto {A^{1/3}} \Rightarrow \frac{{{r_2}}}{{{r_1}}} = {\left( {\frac{{{A_2}}}{{{A_1}}}} \right)^{1/3}} = {\left( {\frac{{206}}{4}} \right)^{1/3}}$

$\therefore {r_2} = 3{\left( {\frac{{206}}{4}} \right)^{1/3}} = 11.6\;Fermi$

Standard 12
Physics

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