Radius nuclear

Nuclei suitable for fusion must come near each other, where near means something like the nuclear radius of 10" cm. For positively charged nuclei to make such a close approach it requires large head-on velocities, and therefore multimillion-degree Celsius temperature. In contrast, fission can occur at normal temperatures, either spontaneously or triggered by a particle, particularly an uncharged neutron, coming near a fissionable nucleus. 

This theory appears not to involve adjustable parameters (other than the nuclear radius parameters that were taken from the literature). In particular, it was criticized that the calibration approach involved a slope that is too high by about a factor of two. However, in actual calculations with the linear response approach, it was found that the slope of the correlation line between theory and experiment (dependent on the quantum chemical method) is close to 0.5. Thus, it also requires a scaling factor of about 2 in order to reach quantitative agreement with experiment. The standard deviations between the calibration and linear response approaches are comparable thus indicating that the major error in both approaches still stems from errors in the description of the bonding that is responsible for the actual valence shell electron distribution. 

The relative change of the mean-square nuclear radius in going from the excited to the ground state, A r )/ r ), is positive for u. An increase in observed isomer shifts S therefore reflects an increase of the s-electron density at the Ru nucleus caused by either an increase in the number of s-valence electrons or a decrease in the number of shielding electrons, preferentially of d-character. 

The isomer shifts in hafnium Mossbauer isotopes usually are of the order of some percent of the line width. Boolchand et al. [168] observed a relatively large isomer shift of -1-0.19 0.06 mm s between cyclopentadienyl hafnium dichloride (Hf(Cp)2Cl2) and Hf metal. From a comparison with Os(Cp)2 and Os-metal, a value of 5 r ) ( Hf) = —0.37 10 fm has been derived, which implies a shrinking of the nuclear radius in the excited 2 state. Figure 7.37 shows some typical spectra for Hf in various hafnium compounds (from [168]). 

For typical lepton energies of a few MeV, the de Broglie wavelength is of order 100 times the nuclear radius and when orbital angular momentum is zero, one can use the allowed approximation for their wave functions... 

The Mossbauer effect involves the resonance fluorescence of nuclear gamma radiation and can be observed during recoilless emission and absorption of radiation in solids. It can be exploited as a spectroscopic method by observing chemically dependent hyperfine interactions. The recent determination of the nuclear radius term in the isomer shift equation for shows that the isomer shift becomes more positive with increasing s electron density at the nucleus. Detailed studies of the temperature dependence of the recoil-free fraction in and labeled Sn/ show that the characteristic Mossbauer temperatures Om, are different for the two atoms. These results are typical of the kind of chemical information which can be obtained from Mossbauer spectra. 

Next to Fe, Sn is the second best Mossbauer isotope, in the sense that the y-ray energy, Mossbauer lifetime, and parent lifetime are all satisfactory. To use isomer shift measurements eflFectively, one must know the relative change in nuclear radius (8R/R). In the case of Sn, the correct value for this quantity was indefinite, both as to sign and magnitude. 

To get a fairly good number for the change in nuclear radius, we can consider a wider class of tin-containing materials. The results in Figure 1 are the work of Lees (12). Here we have taken arbitrarily as our origin, the source, the intermetallic compound Mg2Sn, which provides a convenient cubic environment for tin. It gives a line which has natural... 

Because the interactions measured in Mossbauer experiments are products of atomic and nuclear factors, experiments on iodine isotopes have yielded values of the change of nuclear radius between the ground state and the excited state, AR/R, quadrupole moment values Q, and magnetic moment values, fi, as well as electric field gradients and internal magnetic fields. 

TABLE 3. Nuclear finite-size correction to the energy (in cm ) for the low transitions of Li-like ions, and values of the effective nuclear radius (in 10 cm). 

The main difference between the quoted papers lies in the modeling of the magnetic moment distribution in the nucleus a bulk distribution is assumed in the present paper and in paper [11] and a surface distribution is adopted in ref [22]. A systematic 1% difference is observed, which cannot be explained by the uncertainty in the nuclear radius. However, it is known that variations of the nuclear size within reasonable limits can lead to variations in the value of A of several orders of magnitude [11, 14]. This question will be analysed in a separ-ate paper. 

One can parameterize this distribution by saying that the nuclear radius R can be written as... 

Consider the nuclei 15C, 15N, and 150. Which of these nuclei is stable What types of radioactive decay would the other two undergo Calculate the binding energy difference between 15N and 150. Assuming this difference comes from the Coulomb term in the semiempirical binding energy equation, calculate the nuclear radius. 

The observed half-life of 238U is 4.47 x 109 y, which is a factor of 25 times longer than the calculated value. Note the qualitative aspects of this calculation. The a particle must hit the border of the parent nucleus 1038 times before it can escape. Also note the extreme sensitivity of this calculation to details of the nuclear radius. A 2% change in R changes A. by a factor of 2. In our example, we approximated R as 7 xh + Ra. In reality, the a particle has not fully separated from the daughter nucleus when they exit the barrier. One can correct for this by approximating R 1.4A1 3. 

The creation of relative angular momentum in 3 decay is even more difficult than that in a decay and causes more severe hindrance for each unit of relative angular momentum. The difficulty is easy to see with a simple calculation. We can write the relative angular momentum for two bodies as the cross product L= r x p where r is the radius of emission and p is the momentum. Taking a typical nuclear radius of 5 fm and a typical 3-decay energy of 1 MeV, we find the maximum of the cross product to... 

In VCK lattice P-modification of titanium of emptiness with radius 0,44 nm almost precisely correspond to nuclear radius of hydrogen 0,41 nm, and free fluctuations of atoms between knots do not occur. Hence, the system is more stable. Therefore, the hydrogen is well dissolved in P-phase (up to 2 %), stabilising it [3],... 

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