We have seen that Q for the overall proton–proton fusion cycle is 26.7 MeV. How can you relate this number to the Q values for the reactions that make up this cycle, as displayed in Fig. 43-11?
The given Q-value 26.7 MeV is the total energy of the proton-proton fusion cycle.
The Q-values for the individual reactions of the proton-proton fusion cycle in figure 43-11 are given.
The Q-value of the fusion process is the amount of energy that is either absorbed or released during the process. Thus, it describes the integration process for a fusion process that involves the binding of two nuclei.
The Q-value of a given full cycle is the sum of all the energy (or Q-value) of the reactions that contribute to the full cycle reaction as follows:
Let, the Q-values of all the given individual reactions be , and . Thus, for the overall proton-proton fusion cycle, the total Q-value can be calculated using equation (i) as follows:
Hence, the given value 26.7 MeV is the total energy of the proton-proton fusion cycle.
Figure 43-15 shows an early proposal for a hydrogen bomb. The fusion fuel is deuterium, . The high temperature and particle density needed for fusion are provided by an atomic bomb “trigger” that involves a or fission fuel arranged to impress an imploding, compressive shock wave on the deuterium. The fusion reaction is
(a) Calculate Q for the fusion reaction. For needed atomic masses, see Problem 42. (b) Calculate the rating (see Problem 16) of the fusion part of the bomb if it contains 500 kg of deuterium, 30.0% of which undergoes fusion.
A nuclear reactor is operating at a certain power level, with its multiplication factor k adjusted to unity. If the control rods are used to reduce the power output of the reactor to 25% of its former value, is the multiplication factor now a little less than unity, substantially less than unity, or still equal to unity?
For overcoming the Coulomb barrier for fusion, methods other than heating the fusible material have been suggested. For example, if you were to use two particle accelerators to accelerate two beams of deuterons directly toward each other so as to collide head-on, (a) what voltage would each accelerator require in order for the colliding deuterons to overcome the Coulomb barrier? (b) Why do you suppose this method is not presently used?
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