In nuclear reactors an example reaction would be Np-239 (nuclear mass defect = 49.3110MeV/c^2) to Pu-239...
In nuclear reactors an example reaction would be Np-239 (nuclear mass defect = 49.3110MeV/c^2) to Pu-239 (nuclear mass defect = 48.5882MeV/c^2) + other products. Calculate the magnitude of the difference in total nuclear binding energy before and after. (Answer in MeV with three sig figs)
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In nuclear reactors an example reaction would be Np-239
(nuclear mass defect = 49.3110MeV/c^2) to Pu-239...
In nuclear reactors an example reaction would be Np-239 (nuclear mass defect = 49.3110 MeV/c^2) to...
In nuclear reactors an example reaction would be Np-239 (nuclear mass defect = 49.3110 MeV/c^2) to Pu-239 (nuclear mass defect = 48.5882 MeV/c^2) + other products. For a mole of these reactions calculate the magnitude in the change in energy. (Answer in GJ with three sig figs)
plutonium-239 (239/94 Pu) is an isotope that can be used as a fuel in nuclear reactors....
plutonium-239 (239/94 Pu) is an isotope that can be used as a fuel in nuclear reactors. when the plutonium nucleus is hit by a nuetron it undergoes fission, and splits into two smaller nuclei. there are many possibilities for what the smaller nuclei might be, but we'll just look at one possible fission reaction where it splits into cerium 148 (148/58 Ce) and krypton 89 (89/36 Kr) a) write an equation for the fission of plutonium 239 into cerium 148...
What is the energy released in this β−β−nuclear reaction 239 93Np→239 94Pu+0−1e932 39Np→942 39Pu+−10e? (The atomic mass of 239Np239Np is 239.052939 u and that of 239Pu239Pu is 239.052163 u) 6. What i...
What is the energy released in this β−β−nuclear reaction 239
93Np→239 94Pu+0−1e932 39Np→942 39Pu+−10e? (The atomic mass of
239Np239Np is 239.052939 u and that of 239Pu239Pu is 239.052163
u)
6. What is the energy released in this β 239.052163 u) nuclear reaction 9Np → 9 Pu + ,e? (The atomic mass of 239 Np is 239.052939 u and that of 239Pu is MeV
6. What is the energy released in this β 239.052163 u) nuclear reaction 9Np → 9 Pu...
Calculating Mass Defect and Nuclear Binding Energy
Calculating Mass Defect and Nuclear Binding Energy Learning Goal: To learn how to calculate the binding energy of a nucleus. The measured masses of nuclei are consistently lower than predicted by the sum of their particles. This discrepancy is called the mass defect, and it can be used to calculate the nuclear binding energy according to Einstein's famous equationΔE = Δmc2 where ΔE is the energy produced, Δm is the mass lost, and c 3.00 x 108 m/s. Nuclear binding energy is the energy holding the...
Calculate the mass defect and the nuclear binding energy per nucleon for Ti-48 (atomic mass =...
Calculate the mass defect and the nuclear binding energy per nucleon for Ti-48 (atomic mass = 47.947947 amu). The mass of a proton is 1.00728 amu, the mass of a neutron is 1.008665 amu, and the mass of an electron is 0.00055 amu. A. Nuclear binding energy per nucleon = 5.6062 MeV/nucleon B. Nuclear binding energy per nucleon = 7.0754 MeV/nucleon C. Nuclear binding energy per nucleon = 8.0534 MeV/nucleon D. Nuclear binding energy per nucleon = 8.7204 MeV/nucleon E....
mass particle mass 239 94 Pu 239.0522 u on 1.0086649 u 0 10 -4 5.4858 x...
mass particle mass 239 94 Pu 239.0522 u on 1.0086649 u 0 10 -4 5.4858 x 10 u He 4.00260 297 Cm 242.0588 u 0 -4 5.4858 x 10 u Now consider the following nuclear reaction: He 29Cm + on 239 + -. O released absorbed x ? is energy released or absorbed by this reaction? neither released nor absorbed O I need more information to decide If you said energy was released or absorbed, calculate how much energy is...
Complete the following nuclear equations. 239 Pu 11. 42 K → 19 0e + 16. 94...
Complete the following nuclear equations. 239 Pu 11. 42 K → 19 0e + 16. 94 4 He + 12. Be OY + 17. 32SU + go 142 13. 6 Li He + 18. 56 Ba + 3.Kr +3 ) n 3 14. Calculate the amount of energy released (in kJ/mol) when 235U reacts with one neutron form Te-137 and Zr-96 and 3 neutrons in a fission reaction. 235U = 235.043915 amu In = 1.008664 amu 137 Te = 136.925449...
Nuclear Physics - Alpha Decay a - decay After Before a Parent Daughter 4P→ D +...
Nuclear Physics - Alpha Decay a - decay After Before a Parent Daughter 4P→ D + He Z A plutonium Pu nucleus (94 protons, 239 nucleons) go through an a - decay. (a) Find the number of protons and the number of nucleons in the Daughter nucleus. Enter the number of protons in the first box, and number of nucleons in the 2nd box. Enter a number Submit (5 attempts remaining) Enter a number Submit (5 attempts remaining) The relevant...
5. In a nuclear reaction, the total rest mass of the products is 0.004710 u greater...
5. In a nuclear reaction, the total rest mass of the products is 0.004710 u greater than the total rest (1 point) mass of the reactants. Is the reaction exoergic or endoergic, and what is the reaction energy Q? Let lu=931.5 MeV. endoergic; Q=-4.387 MeV exoergic; Q=4.387 MeV endoergic; Q=4.710 MeV exoergic; Q = -1.753 MeV 7. The uranium used in nuclear reactors is "enriched." What is done to the uranium to enrich it? (1 point) The external pressure on...
Problem 5. [Einstein's E = mc²] Sun is producing energy by nuclear fusion reactions converting four...
Problem 5. [Einstein's E = mc²] Sun is producing energy by nuclear fusion reactions converting four hydrogen nuclei to form one nucleus of He, i.e. 4H + He. However, mass of helium nucleus is smaller than that of four hydrogens and this difference is called mass defect. If mass of the hydrogen nucleus is my and mass of the helium nucleus Mhe mass defect of one nuclear reaction Am is equal to Am = 4 *my - Me (1) Using...
What is the energy released in this β−β−nuclear reaction 239
93Np→239 94Pu+0−1e932 39Np→942 39Pu+−10e? (The atomic mass of
239Np239Np is 239.052939 u and that of 239Pu239Pu is 239.052163
u)
6. What is the energy released in this β 239.052163 u) nuclear reaction 9Np → 9 Pu + ,e? (The atomic mass of 239 Np is 239.052939 u and that of 239Pu is MeV
6. What is the energy released in this β 239.052163 u) nuclear reaction 9Np → 9 Pu...
mass particle mass 239 94 Pu 239.0522 u on 1.0086649 u 0 10 -4 5.4858 x 10 u He 4.00260 297 Cm 242.0588 u 0 -4 5.4858 x 10 u Now consider the following nuclear reaction: He 29Cm + on 239 + -. O released absorbed x ? is energy released or absorbed by this reaction? neither released nor absorbed O I need more information to decide If you said energy was released or absorbed, calculate how much energy is...
Complete the following nuclear equations. 239 Pu 11. 42 K → 19 0e + 16. 94 4 He + 12. Be OY + 17. 32SU + go 142 13. 6 Li He + 18. 56 Ba + 3.Kr +3 ) n 3 14. Calculate the amount of energy released (in kJ/mol) when 235U reacts with one neutron form Te-137 and Zr-96 and 3 neutrons in a fission reaction. 235U = 235.043915 amu In = 1.008664 amu 137 Te = 136.925449...
Nuclear Physics - Alpha Decay a - decay After Before a Parent Daughter 4P→ D + He Z A plutonium Pu nucleus (94 protons, 239 nucleons) go through an a - decay. (a) Find the number of protons and the number of nucleons in the Daughter nucleus. Enter the number of protons in the first box, and number of nucleons in the 2nd box. Enter a number Submit (5 attempts remaining) Enter a number Submit (5 attempts remaining) The relevant...
5. In a nuclear reaction, the total rest mass of the products is 0.004710 u greater than the total rest (1 point) mass of the reactants. Is the reaction exoergic or endoergic, and what is the reaction energy Q? Let lu=931.5 MeV. endoergic; Q=-4.387 MeV exoergic; Q=4.387 MeV endoergic; Q=4.710 MeV exoergic; Q = -1.753 MeV 7. The uranium used in nuclear reactors is "enriched." What is done to the uranium to enrich it? (1 point) The external pressure on...
Problem 5. [Einstein's E = mc²] Sun is producing energy by nuclear fusion reactions converting four hydrogen nuclei to form one nucleus of He, i.e. 4H + He. However, mass of helium nucleus is smaller than that of four hydrogens and this difference is called mass defect. If mass of the hydrogen nucleus is my and mass of the helium nucleus Mhe mass defect of one nuclear reaction Am is equal to Am = 4 *my - Me (1) Using...
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