Respuesta :
Answer:
0.22 b
Explanation:
Quadrupole moment of the nucleon is,
[tex]Q=-\frac{2j-1}{2(j+1)}\frac{3}{5}R^{2}[/tex]
And also,
[tex]R^{2}=R^{2} _{0}A^{\frac{2}{3} }[/tex]
And, [tex]R _{0}=1.2\times 10^{-15}m[/tex]
Now,
[tex]Q=-\frac{2j-1}{2(j+1)}\frac{3}{5}R^{2} _{0}A^{\frac{2}{3} }[/tex]
For Bismuth [tex]j=\frac{9}{2}[/tex] and A is 209.
[tex]Q=-\frac{2\frac{9}{2} -1}{2(\frac{9}{2} +1)}\frac{3}{5}(1.2\times 10^{-15}) ^{2}(209)^{\frac{2}{3} }\\Q=0.628\times 35.28\times 10^{-30} \\Q=22.15\times 10^{-30} m^{2} \\Q=0.2215\times 10^{-28} m^{2} \\Q=0.22 barn[/tex]
Therefore, the expected value of quadrupole is 0.22 b which is quite related with experimental value which is 0.37 b
