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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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Answer:

bromine

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3 years ago
What information can scientists obtian from tree rings
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3 years ago
Imagine that you are working as a roller coaster designer. You want to build a record breaking coaster that goes 70.0 m/s at the
Rzqust [24]

Wow !  This is not simple.  At first, it looks like there's not enough information, because we don't know the mass of the cars.  But I"m pretty sure it turns out that we don't need to know it.

At the top of the first hill, the car's potential energy is

                                  PE = (mass) x (gravity) x (height) .

At the bottom, the car's kinetic energy is

                                 KE = (1/2) (mass) (speed²) .

You said that the car's speed is 70 m/s at the bottom of the hill,
and you also said that 10% of the energy will be lost on the way
down.  So now, here comes the big jump.  Put a comment under
my answer if you don't see where I got this equation:

                                   KE = 0.9  PE

        (1/2) (mass) (70 m/s)² = (0.9) (mass) (gravity) (height)     

Divide each side by (mass): 

               (0.5) (4900 m²/s²) = (0.9) (9.8 m/s²) (height)

(There goes the mass.  As long as the whole thing is 90% efficient,
the solution will be the same for any number of cars, loaded with
any number of passengers.)

Divide each side by (0.9):

               (0.5/0.9) (4900 m²/s²) = (9.8 m/s²) (height)

Divide each side by (9.8 m/s²):

               Height = (5/9)(4900 m²/s²) / (9.8 m/s²)

                          =  (5 x 4900 m²/s²) / (9 x 9.8 m/s²)

                          =  (24,500 / 88.2)  (m²/s²) / (m/s²)

                          =        277-7/9    meters
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3 0
3 years ago
According to Newton’s first law of motion, when will an object at rest begin to move?
Natasha_Volkova [10]

Newton's first law of motion says something like "An object remains
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8 0
3 years ago
Read 2 more answers
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kow [346]

Answer:

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Explanation:

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Initial quantity of the sample: 2.5 grams.

After 28 years, the leftover quantity = 1.25 grams

After 56 years, the leftover quantity = 0.625 grams

After 84 Years, the leftover quantity = 0.3125 grams

After 112 years, the leftover quantity = 0.15625 grams

5 0
3 years ago
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