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Naya [18.7K]
3 years ago
8

In odd nuclei, what determines the final spin of the nucleus?

Physics
1 answer:
Katen [24]3 years ago
7 0

Answer:

In odd nuclei, the left out proton or neutron will contribute to the spin of the nucleus.

Explanation:

The meaning of odd nuclei is atomic mass is odd.

A=odd number.

A=Z+n

Here, Z is proton either it will odd or n will odd which is neutron.

Now according to the shell model the left out proton or neutron will contribute to the spin and parity.

For example,

Take the case of isotope of nitrogen-15.

Here Z is 7, and n is 8 will not contribute in spin.

Now, for Z=7.

1S^{2} _{\frac{1}{2} }, 1P^{4} _{\frac{3}{2} }, 1P^{1} _{\frac{1}{2} }

Here,

j=\frac{1}{2}

and, L=1.

Fort parity,

(-1)^{L}

Put the value of L.

Parity will be -1.

Now, spin will be

S=(\frac{1}{2} )^{-1}.

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U= \frac{1}{2}kx^2 =  \frac{1}{2}(35.0 N/m)(0.01 m)^2 = 1.75 \cdot 10^{-3} J
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K=E-U=0.028 J - 1.75 \cdot 10^{-3}J =0.026 J
since the mass is m=50.0 g=0.05 kg, and the kinetic energy is given by
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we can re-arrange the formula to find the speed of the object:
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Answer:

(a) a = 1.875 m/s²

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m : mass s (kg)

a : acceleration  (m/s²)

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WA: Weight of of the A block : In vertical direction  downaward (-y)

NA : Normal force of the A block :In vertical direction  upaward (+y)

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WB: Weight of the B block: In vertical direction  downaward (-y)

NB : Normal force of the B block :In vertical direction  upaward (+y)

T :  tension of the string: In in the direction (+x)

Data

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mB = 0.84 kg

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a =5.7 / (3.04)

a = 1.875 m/s²

(b)Tension (in N) in the string connecting the two blocks

We replace data in the  Equation (1)

T = (0.84)*a

T = (0.84)*(1.875)

T = 1.575 N

(c) How will the tension in the string be affected if mA is decreased?

We observed Equation (2)

5.7 = (2.2)*a + (0.84)*a

5.7 = (mA)*a + (0.84)*a

5.7 =a*( mA+ 0.84)

if mA decrease , then, the acceleration increase and T  increase

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