High temperatures are required for controlled nuclear fusion because nuclei must O produce antimatter. O have enough energy to oppose repulsive forces in the nucleus. O undergo a state change from solid to liquid in order to combine. O have enough energy to overcome the gravitational attraction of nuclei.

What minimum energy E(min) is needed to remove a neutron from K40 and so convert it to K39? The atomic masses of the two isotopes are 39.964000 and 38.963708 u, respectively. E(min) = ? eV

Consider the following a-decay of the Uranium nucleus 236 U → 332Th + a. 90 (a) Show how the mass number (Aa) and atomic number (Za) of the alpha particle are obtained from this equation. (b) Calculate the Q-value (Qa) of the reaction. (c) Calculate the speed va = √2MQ of the alpha particle after it has been ejected from the parent nucleus, in terms of the speed of light c. M = -, mp and ma are the atomic masses (ma+mp) x ma mp of the daugher nucleus and the alpha particle, respectively. (d) Calculate the classical turning radius Re= 2Zpe²/Qa, where Zp is the atomic number of the daughter nucleus, and e² = 1.44 MeV. fm. (e) Calculate the decay probability Pa that the alpha particle will tunnel through the barrier. HINT: In calculating the probability, use the fact that ħc=197.327 MeV fm. The formula to use is given on page 5.

One of the ways that 235U can fission is by means of the following reaction: In +235 U 141 Ba + Kr + 3 n 56 Atomic masses of the reactants and products are as follows: n 235 U 92 92 Kr 36 141 Ba 56 1.008665 u 235.043930 u 91.926156 u 140.9144035 u a) Determine the energy in MeV that is released when one 23592U nucleus fissions. Assume that the incoming neutron is very slow. b) Find the total energy released in MeV if 2.8 kg of 23592U were to undergo fission entirely by this reaction.

A new element has been created artificially. It is named Elementy McFacium. The most common isotope of McFacium has 153 protons and 277 neutrons. The mass of this isotope of the new element is 402,188.91 MeV/c^2. What is the total mass deficit for the nucleus?  Hint: The new element is not made as an ion! Use the mass of hydrogen atoms instead of protons.          (Answer in atomic mass units u)

1) Complete the reaction: 242Am +      ?      n → 90Sr + 149La + 4n a) n b) 2n c) 3n d) 4n  e) 5n 2) Calculate the energy (in MeV) released in the nuclear fission reaction. The atomic masses are 242Am = 242.059549 u, 90Sr = 89.9077387 u, and 149La = 148.934733 u. (Express your answer to six significant figures).

A "mad" scientist attempts to take a sample of Mercury-199 and bombard it with neutrons. The reaction creates a proton and Gold-199. His joy is short-lived when in a few days the radioactive Gold-199 reverts back to Mercury-199 emitting a single particle. What is the emitted particle? 199 79 O O → 199 80Hg+? a beta (-) particle ß- a proton; H+ an alpha particle; a a neutron; nº

Ordinary hydrogen is sometimes called a perfect fuel, both because of its almost unlimited supply on Earth and because, when it burns, harmless water is the product of the combustion. So why don’t we abandon fission and fusion energies, not to mention fossil-fuel energy, and just use hydrogen?

The indium isotope 115/49 I n captures an electron and becomes the cadmium isotope 115/48 C d, as shown in the nuclear reaction below. 115/49 I n + ? ⟶ 115/48 C d What kind of particle is emitted in this decay, as denoted by the question mark?