1.a radioscope which is neutron poor and very heavy is most likely to decay by?
alpha , beta, electron capture, positron emission
2. When a 235-U nucleus is struck by a thermal neutron, fission occurs often with the release of neutrons. If fission fragments are 89-Sr and 144-Xe, how many neutrons are released?
Alpha emission, electron capture, and positon emission are possible for neutron poor radioisotopes. The most likely is alpha.
For part 2, first calculate the number of neutrons in 89Sr and 144Xe. How do you do that?
Look up the atomic number for Sr. That is 38. The number of neutrons, then, is 89-38=??
Do the same thing for Xe. Add the neutrons together.
Now determine the number of nuetrons in 235U which is 235-92. The difference between what was available in U and what was obtained (Sr+Xe) is the number of neutrons that were released. Post your work if you get stuck.
1) alpha emission
1. In the given scenario, where a radioscope is neutron-poor and very heavy, the most likely decay mode is alpha decay. This is because alpha decay occurs in heavy nuclei that have an excess of protons. The decay process involves the emission of an alpha particle, which consists of two protons and two neutrons.
2. To determine the number of neutrons released in the fission of a 235-U nucleus, we need to calculate the difference between the number of neutrons in the fission fragments (89-Sr and 144-Xe) and the number of neutrons in the original 235-U nucleus.
First, we find the number of neutrons in 89-Sr. Since the atomic number of strontium (Sr) is 38, subtracting it from the mass number (89) gives us 89 - 38 = 51 neutrons.
Next, we determine the number of neutrons in 144-Xe. With the atomic number of xenon (Xe) being 54, subtracting it from the mass number (144) gives us 144 - 54 = 90 neutrons.
Now, we calculate the number of neutrons in the 235-U nucleus. The mass number (235) minus the atomic number (92) gives us 235 - 92 = 143 neutrons.
Finally, we subtract the total number of neutrons in the fission fragments (51 + 90 = 141) from the number of neutrons in the 235-U nucleus (143). Therefore, the number of neutrons released in this fission reaction is 143 - 141 = 2 neutrons.