15. what are the pros and cons of nuclear energy?\n16. determine the other product of the following decay…
15. what are the pros and cons of nuclear energy?\n16. determine the other product of the following decay examples:\n a. ^{242}cm → ^{4}he + \n b. ^{63}ni → ^{0}e + \n c. what type of decay is shown in a?\n d. what type of decay is shown in b?\n17. what is the ratio of protons to neutrons that is in the stable range?\n18. which elements are always unstable?
Answer
15.
Brief Explanations:
Pros: Nuclear energy is a low - carbon energy source, providing a large amount of baseload power. It has high energy density, meaning a small amount of fuel can produce a large amount of energy. Cons: There are concerns about radioactive waste disposal, which remains hazardous for long periods. Nuclear accidents, although rare, can have catastrophic consequences for human health and the environment. Also, nuclear fuel proliferation is a security risk.
Answer:
Pros: Low - carbon, high energy density. Cons: Radioactive waste, accident risks, proliferation concerns.
16.
Explanation:
Step1: For part a, use conservation of mass - number and atomic number
In a nuclear decay, the sum of mass - numbers and atomic numbers on both sides of the equation must be equal. The atomic number of $Cm$ (Curium) is 96 and its mass - number is 242, and for $He$ (Helium) the atomic number is 2 and mass - number is 4. Let the unknown product have atomic number $Z$ and mass - number $A$. Then $A=242 - 4=238$ and $Z = 96-2 = 94$. The element with atomic number 94 is $Pu$ (Plutonium). So the product is $^{238}Pu$.
Step2: For part b, use conservation of mass - number and atomic number
The atomic number of $Ni$ (Nickel) is 28 and mass - number is 63, and for $e$ (electron) the atomic number is - 1 and mass - number is 0. Let the unknown product have atomic number $Z$ and mass - number $A$. Then $A=63-0 = 63$ and $Z=28+1=29$. The element with atomic number 29 is $Cu$ (Copper). So the product is $^{63}Cu$.
Step3: Identify the decay type in part a
The decay in part a is an alpha - decay because an alpha particle ($^{4}He$) is emitted.
Step4: Identify the decay type in part b
The decay in part b is a beta - decay because an electron ($^{0}_{- 1}e$) is emitted.
Answer:
a. $^{238}Pu$ b. $^{63}Cu$ c. Alpha - decay d. Beta - decay
17.
Brief Explanations:
For light elements (up to about $Z = 20$), the stable proton - neutron ratio is close to 1:1. For heavier elements, the stable ratio is closer to 1.5:1 as the number of neutrons needs to be greater than the number of protons to counteract the electrostatic repulsion between protons.
Answer:
For light elements: 1:1; for heavier elements: about 1.5:1
18.
Brief Explanations:
Elements with atomic number greater than 82 (Lead) are always unstable because the electrostatic repulsion between protons becomes too great to be balanced by the strong nuclear force, even with a sufficient number of neutrons.
Answer:
Elements with atomic number $Z>82$.