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Studentka2010 [4]
3 years ago
5

Compute the expected shell-model quadrupole moment of 209Bi () and compare with the experimental value, - 0.37 b

Physics
1 answer:
Over [174]3 years ago
3 0

Answer:

0.22 b

Explanation:

Quadrupole moment of the nucleon is,

Q=-\frac{2j-1}{2(j+1)}\frac{3}{5}R^{2}

And also,

R^{2}=R^{2} _{0}A^{\frac{2}{3} }

And, R _{0}=1.2\times 10^{-15}m

Now,

Q=-\frac{2j-1}{2(j+1)}\frac{3}{5}R^{2} _{0}A^{\frac{2}{3} }

For Bismuth j=\frac{9}{2} and A is 209.

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

Therefore, the expected value of quadrupole is 0.22 b which is quite related with experimental value which is 0.37 b

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\boxed{\sf Velocity=\dfrac{Displacement}{Time}}

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Explanation:

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Answer:

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The heat of reaction or enthalpy of reaction (ΔH) is the amount of heat energy that the system must release or absorb so that the temperature remains constant throughout the chemical reaction process. In other words, the heat of reaction or enthalpy of reaction (ΔH) is the change in the enthalpy of a chemical reaction (the energy  absorbed or released into it) that occurs at a constant pressure.

Then, this energy can be observed in the following way:

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