The minimum kinetic energy needed by an alpha particle to cause the nuclear reaction

\( \newcommand\): Alpha decay schematics (CC BY-NC-ND; Paola Cappellaro) Energetics In analyzing a radioactive decay (or any nuclear reaction) an important quantity is \(Q\), the net energy released in the decay: \(Q=\left(m_\) Quantum mechanics description of alpha decay We will use a semi-classical model (that is, combining quantum mechanics with classical physics) to answer the questions above. In order to study the quantum mechanical process underlying alpha decay, we consider the interaction between the daughter nuclide and the alpha particle. Just prior to separation, we can consider this pair to be already present inside the parent nuclide, in a bound state. We will describe this pair of particles in their center of mass coordinate frames: thus we are interested in the relative motion (and kinetic energy) of the two particles. As often done in these situations, we can describe the relative motion of two particles as the motion of a single particle of reduced mass \(\mu=\frac \) but also on Z 1Z 2 (where Z i are the number of protons in the two daughters). This leads to the following observations: • Other types of decay are less likely, because the Coulomb energy would increase considerably, thus the barrier becomes too high to be overcome. • The same is true for spontaneous fission, despite the fact that \(Q\) is much higher (∼ 200MeV). • We thus find that alpha decay is the optimal mechanism. Still, it can happen only for A ≥ 200 exactly because otherwise the tunne...

The minimum kinetic energy needed by an alpha particle to cause the nuclear reaction 716 N + 24 He arrow 11 H + 819 O in a laboratory frame is n (in MeV ). Assume that 716 N is at rest in the laboratory frame. The masses of 716 N , 24 He , 11 H and 819 O can be taken to be 16.006 u, 4.003 u, 1.008 u and 19.003 u, respectively, where 1 u=930 MeVc -2. The value of n is Q. The minimum kinetic energy needed by an alpha particle to cause the nuclear reaction 7 16 ​ N + 2 4 ​ He → 1 1 ​ H + 8 19 ​ O in a laboratory frame is n (in M e V ). Assume that 7 16 ​ N is at rest in the laboratory frame. The masses of 7 16 ​ N , 2 4 ​ He , 1 1 ​ H and 8 19 ​ O can be taken to be 16.006 u , 4.003 u, 1.008 u and 19.003 u, respectively, where 1 u = 930 M e V c − 2. The value of n is ___ Solution: 7 16 ​ N + 2 4 ​ He → 1 1 ​ He + 8 19 ​ O 7 16 ​ N + 2 4 ​ He → 1 1 ​ He + 8 19 ​ O 16.006 4.003 1.008 19.003 4 v 0 ​ = 1 v 1 ​ + 19 v 2 ​ = 20 v 2 ​ (For max loss of K E ) v 0 ​ = 5 v 2 ​ ​ E required = ( 1.008 + 19.003 − 16.006 − 4.003 ) × 930 = 1.86 2 1 ​ 4 v 0 2 ​ − 2 1 ​ 20 v 2 = 1.86 2 1 ​ 4 v 0 2 ​ − 10 25 v 0 2 ​ ​ 20 v 2 = 1.86 2 v 0 2 ​ − 5 2 ​ v 0 2 ​ = 1.86 5 8 ​ v 0 2 ​ = 1.86 v 0 2 ​ = 8 1.86 × 5 ​ K E = 2 1 ​ 4 v 0 2 ​ = 2 v 0 2 ​ = 4 18.6 × 5 ​ = 2.325

Coulomb Barrier for Nuclear Fusion Coulomb Barrier for Fusion In order to accomplish Considering the barrier to be the where k = Given the radius r at which the nuclear attractive force becomes dominant, the temperature necessary to raise the average R Nave Calculation of Coulomb Barrier The height of the The For A a = and Z a = , R a = x10^ m = fermi. For A b = and Z b = , R b = x10^ m = fermi. The height of the Coulomb Barrier is V C = x 10^ J = x 10^ eV = keV = MeV. The temperature required to provide this energy as an average x 10^ K Of course for head-on collisions between particles only half of that energy would be required of each particle, so you could cut that temperature in half. The above temperature is calculated as a reference value. References: Sec 14.2 R Nave Critical Ignition Temperature for Fusion The The presumed height of the coulomb barrier is based upon the distance at which the When trying to model the probability of nuclear fusion, the typical approach is to model it as a " With all these considerations, the energy requirements for the nuclear fusion of deuterium- tritium and for deuterium-deuterium are dramatically lower than the coulomb barrier, but they are nevertheless very difficult to attain in a controlled manner. The references suggest that the currently attainable particle energies in thermonuclear reactors is in the range 1 - 10 keV. If you just substitute those energies into the 7K and 10 keV corresponds to 0.77 x 10 8K. The 8 K, well abov...

Since, the masses are given in atomic mass unit, it is easiest to proceed by finding the mass difference between reactants and products in the same units and then multiplying by 931.5 MeV/u. Thus, we have Q value is negative. It means reaction is endothermic. So, the minimum kinetic energy of α-particle to initiate this reaction would be, Categories • • (31.9k) • (8.8k) • (764k) • (261k) • (257k) • (218k) • (248k) • (2.9k) • (5.2k) • (664) • (121k) • (72.1k) • (3.8k) • (19.6k) • (1.4k) • (14.2k) • (12.5k) • (9.3k) • (7.7k) • (3.9k) • (6.7k) • (63.8k) • (26.6k) • (23.7k) • (14.6k) • (25.7k) • (530) • (84) • (766) • (49.1k) • (63.8k) • (1.8k) • (59.3k) • (24.5k)

Learning Objectives By the end of this section, you will be able to: • Describe and compare three types of nuclear radiation • Use nuclear symbols to describe changes that occur during nuclear reactions • Describe processes involved in the decay series of heavy elements Early experiments revealed three types of nuclear “rays” or radiation: alpha ( α α) rays, beta ( β β) rays, and gamma ( γ γ) rays. These three types of radiation are differentiated by their ability to penetrate matter. Alpha radiation is barely able to pass through a thin sheet of paper. Beta radiation can penetrate aluminum to a depth of about 3 mm, and gamma radiation can penetrate lead to a depth of 2 or more centimeters ( Figure 10.11 A comparison of the penetration depths of alpha ( α α), beta ( β β), and gamma ( γ γ) radiation through various materials. The electrical properties of these three types of radiation are investigated by passing them through a uniform magnetic field, as shown in F → = q v → × B → , F → = q v → × B → , positively charged particles are deflected upward, negatively charged particles are deflected downward, and particles with no charge pass through the magnetic field undeflected. Eventually, α α rays were identified with helium nuclei ( 4 He ) , β ( 4 He ) , β rays with electrons and positrons (positively charged electrons or antielectrons), and γ γ rays with high-energy photons. We discuss alpha, beta, and gamma radiation in detail in the remainder of this section. Figure 10.1...