Answer :
To address the given problem, we need to determine the moment of inertia of a nucleus in various excited states and compare it to the moment of inertia of the nucleus considered as a rigidly rotating sphere.
### Given Data:
- Spin-parity states ([tex]\(j^p\)[/tex]): [tex]\(0^+,\ 2^+,\ 4^+,\ 6^+,\ 8^+\)[/tex]
- Excitation energy [tex]\(E\)[/tex] in keV: 0,\ 100,\ 321,\ 641,\ 1041
- Nucleus: [tex]\( {}_{72}^{170}Hf \)[/tex]
### Constants:
- Mass number, [tex]\(A = 170\)[/tex]
- Radius of the nucleus [tex]\(R = 1.3 \times 10^{-15} \times A^{1/3} \ m\)[/tex]
- Atomic mass unit in kg, [tex]\(1.66053906660 \times 10^{-27} \ kg\)[/tex]
- Reduced Planck constant, [tex]\( \hbar = 1.054571817 \times 10^{-34} \ J \cdot s \)[/tex]
- 1 eV in Joules, [tex]\(1.60218 \times 10^{-19} \ J\)[/tex]
### Calculation Steps:
1. Calculate the Radius of the Nucleus:
[tex]\[ R = 1.3 \times 10^{-15} \times 170^{1/3} \approx 1.3 \times 10^{-15} \times 5.4848 \approx 7.13024 \times 10^{-15} \ m \][/tex]
2. Calculate the Mass of the Nucleus:
[tex]\[ M = 170 \times 1.66053906660 \times 10^{-27} \approx 2.82291641322 \times 10^{-25} \ kg \][/tex]
3. Calculate the Moment of Inertia of the Nucleus as a Rigidly Rotating Sphere:
Using the formula for the moment of inertia of a sphere,
[tex]\[ I_{\text{rigid}} = \frac{2}{5} M R^2 \][/tex]
[tex]\[ I_{\text{rigid}} = \frac{2}{5} \times 2.82291641322 \times 10^{-25} \times (7.13024 \times 10^{-15})^2 \][/tex]
[tex]\[ I_{\text{rigid}} \approx 5.856129370646507 \times 10^{-54} \ kg \cdot m^2 \][/tex]
4. Calculate the Moment of Inertia for Each Excited State:
The energy differences ([tex]\(E_{i+1} - E_i\)[/tex]) for the states are: 100 keV, 221 keV, 320 keV, and 400 keV.
Using the rotational energy formula,
[tex]\[ E = \frac{\hbar^2}{2I} j (j + 1) \][/tex]
and the difference between the levels ([tex]\(\Delta E = E_{i+1} - E_i\)[/tex]),
[tex]\[ \Delta E \left(\text{in Joules}\right) = \text{energy difference in keV} \times 1.60218 \times 10^{-16} \][/tex]
[tex]\[ \Delta j(j + 1) = j_{i+1}(j_{i+1} + 1) - j_i(j_i + 1) \][/tex]
- For [tex]\(j=2^+\)[/tex] to [tex]\(j=0^+\)[/tex]:
[tex]\[ \Delta j(j + 1) = 2(2 + 1) - 0(0 + 1) = 6 \][/tex]
[tex]\[ I = \frac{2 \times 100 \times 1.60218 \times 10^{-16}}{6 \times (1.054571817 \times 10^{-34})^2} \][/tex]
[tex]\[ I \approx 4.80217220592975 \times 10^{56} \ kg \cdot m^2 \][/tex]
- For [tex]\(j=4^+\)[/tex] to [tex]\(j=2^+\)[/tex]:
[tex]\[ \Delta j(j + 1) = 4(4 + 1) - 2(2 + 1) = 12 \][/tex]
[tex]\[ I = \frac{2 \times 221 \times 1.60218 \times 10^{-16}}{12 \times (1.054571817 \times 10^{-34})^2} \][/tex]
[tex]\[ I \approx 4.54834310361632 \times 10^{56} \ kg \cdot m^2 \][/tex]
- For [tex]\(j=6^+\)[/tex] to [tex]\(j=4^+\)[/tex]:
[tex]\[ \Delta j(j + 1) = 6(6 + 1) - 4(4 + 1) = 20 \][/tex]
[tex]\[ I = \frac{2 \times 320 \times 1.60218 \times 10^{-16}}{20 \times (1.054571817 \times 10^{-34})^2} \][/tex]
[tex]\[ I \approx 4.190986652447782 \times 10^{56} \ kg \cdot m^2 \][/tex]
- For [tex]\(j=8^+\)[/tex] to [tex]\(j=6^+\)[/tex]:
[tex]\[ \Delta j(j + 1) = 8(8 + 1) - 6(6 + 1) = 30 \][/tex]
[tex]\[ I = \frac{2 \times 400 \times 1.60218 \times 10^{-16}}{30 \times (1.054571817 \times 10^{-34})^2} \][/tex]
[tex]\[ I \approx 3.8417377647438005 \times 10^{56} \ kg \cdot m^2 \][/tex]
### Summary of Moments of Inertia:
- I_states: [tex]\([0, 4.80217220592975 \times 10^{56}, 4.54834310361632 \times 10^{56}, 4.190986652447782 \times 10^{56}, 3.8417377647438005 \times 10^{56}] \ kg \cdot m^2\)[/tex]
- [tex]\(I_{\text{rigid}} \approx 5.856129370646507 \times 10^{-54} \ kg \cdot m^2\)[/tex]
### Comment on Results:
The calculated moments of inertia for the excited states are orders of magnitude larger than the moment of inertia of the rigidly rotating sphere. This disparity suggests that the simple rigid-body rotation model does not adequately describe the internal dynamics of the nucleus. Various internal nuclear interactions and collective motions contribute to the observed rotational states, and a more complex model is appropriate to describe nuclear structure and its excited states.
### Given Data:
- Spin-parity states ([tex]\(j^p\)[/tex]): [tex]\(0^+,\ 2^+,\ 4^+,\ 6^+,\ 8^+\)[/tex]
- Excitation energy [tex]\(E\)[/tex] in keV: 0,\ 100,\ 321,\ 641,\ 1041
- Nucleus: [tex]\( {}_{72}^{170}Hf \)[/tex]
### Constants:
- Mass number, [tex]\(A = 170\)[/tex]
- Radius of the nucleus [tex]\(R = 1.3 \times 10^{-15} \times A^{1/3} \ m\)[/tex]
- Atomic mass unit in kg, [tex]\(1.66053906660 \times 10^{-27} \ kg\)[/tex]
- Reduced Planck constant, [tex]\( \hbar = 1.054571817 \times 10^{-34} \ J \cdot s \)[/tex]
- 1 eV in Joules, [tex]\(1.60218 \times 10^{-19} \ J\)[/tex]
### Calculation Steps:
1. Calculate the Radius of the Nucleus:
[tex]\[ R = 1.3 \times 10^{-15} \times 170^{1/3} \approx 1.3 \times 10^{-15} \times 5.4848 \approx 7.13024 \times 10^{-15} \ m \][/tex]
2. Calculate the Mass of the Nucleus:
[tex]\[ M = 170 \times 1.66053906660 \times 10^{-27} \approx 2.82291641322 \times 10^{-25} \ kg \][/tex]
3. Calculate the Moment of Inertia of the Nucleus as a Rigidly Rotating Sphere:
Using the formula for the moment of inertia of a sphere,
[tex]\[ I_{\text{rigid}} = \frac{2}{5} M R^2 \][/tex]
[tex]\[ I_{\text{rigid}} = \frac{2}{5} \times 2.82291641322 \times 10^{-25} \times (7.13024 \times 10^{-15})^2 \][/tex]
[tex]\[ I_{\text{rigid}} \approx 5.856129370646507 \times 10^{-54} \ kg \cdot m^2 \][/tex]
4. Calculate the Moment of Inertia for Each Excited State:
The energy differences ([tex]\(E_{i+1} - E_i\)[/tex]) for the states are: 100 keV, 221 keV, 320 keV, and 400 keV.
Using the rotational energy formula,
[tex]\[ E = \frac{\hbar^2}{2I} j (j + 1) \][/tex]
and the difference between the levels ([tex]\(\Delta E = E_{i+1} - E_i\)[/tex]),
[tex]\[ \Delta E \left(\text{in Joules}\right) = \text{energy difference in keV} \times 1.60218 \times 10^{-16} \][/tex]
[tex]\[ \Delta j(j + 1) = j_{i+1}(j_{i+1} + 1) - j_i(j_i + 1) \][/tex]
- For [tex]\(j=2^+\)[/tex] to [tex]\(j=0^+\)[/tex]:
[tex]\[ \Delta j(j + 1) = 2(2 + 1) - 0(0 + 1) = 6 \][/tex]
[tex]\[ I = \frac{2 \times 100 \times 1.60218 \times 10^{-16}}{6 \times (1.054571817 \times 10^{-34})^2} \][/tex]
[tex]\[ I \approx 4.80217220592975 \times 10^{56} \ kg \cdot m^2 \][/tex]
- For [tex]\(j=4^+\)[/tex] to [tex]\(j=2^+\)[/tex]:
[tex]\[ \Delta j(j + 1) = 4(4 + 1) - 2(2 + 1) = 12 \][/tex]
[tex]\[ I = \frac{2 \times 221 \times 1.60218 \times 10^{-16}}{12 \times (1.054571817 \times 10^{-34})^2} \][/tex]
[tex]\[ I \approx 4.54834310361632 \times 10^{56} \ kg \cdot m^2 \][/tex]
- For [tex]\(j=6^+\)[/tex] to [tex]\(j=4^+\)[/tex]:
[tex]\[ \Delta j(j + 1) = 6(6 + 1) - 4(4 + 1) = 20 \][/tex]
[tex]\[ I = \frac{2 \times 320 \times 1.60218 \times 10^{-16}}{20 \times (1.054571817 \times 10^{-34})^2} \][/tex]
[tex]\[ I \approx 4.190986652447782 \times 10^{56} \ kg \cdot m^2 \][/tex]
- For [tex]\(j=8^+\)[/tex] to [tex]\(j=6^+\)[/tex]:
[tex]\[ \Delta j(j + 1) = 8(8 + 1) - 6(6 + 1) = 30 \][/tex]
[tex]\[ I = \frac{2 \times 400 \times 1.60218 \times 10^{-16}}{30 \times (1.054571817 \times 10^{-34})^2} \][/tex]
[tex]\[ I \approx 3.8417377647438005 \times 10^{56} \ kg \cdot m^2 \][/tex]
### Summary of Moments of Inertia:
- I_states: [tex]\([0, 4.80217220592975 \times 10^{56}, 4.54834310361632 \times 10^{56}, 4.190986652447782 \times 10^{56}, 3.8417377647438005 \times 10^{56}] \ kg \cdot m^2\)[/tex]
- [tex]\(I_{\text{rigid}} \approx 5.856129370646507 \times 10^{-54} \ kg \cdot m^2\)[/tex]
### Comment on Results:
The calculated moments of inertia for the excited states are orders of magnitude larger than the moment of inertia of the rigidly rotating sphere. This disparity suggests that the simple rigid-body rotation model does not adequately describe the internal dynamics of the nucleus. Various internal nuclear interactions and collective motions contribute to the observed rotational states, and a more complex model is appropriate to describe nuclear structure and its excited states.
