Learning Materials
Features
Discover
Chapter 31: Q19 E (page 1149)
\({{\rm{\beta }}^{\rm{ + }}}\)decay of \(^{{\rm{50}}}{\rm{Mn}}\)
The \({\beta ^ - }\) Decay equation of \(^{50}Mn\) is \(_{25}^{50}M{n_{25}} \to _{24}^{50}C{r_{26}} + {\beta ^ + } + {\nu _c}\).
Over 30 million students worldwide already upgrade their learning with 91影视!
All the tools & learning materials you need for study success - in one app.
Get started for free
Radioactivity depends on the nucleus and not the atom or its chemical state. Why, then, is one kilogram of uranium more radioactive than one kilogram of uranium hexafluoride?
Electron capture decay equation of 106In
Ionizing radiation interacts with matter by scattering from electrons and nuclei in the substance. Based on the law of conservation of momentum and energy, explain why electrons tend to absorb more energy than nuclei in these interactions.
The relatively scarce naturally occurring calcium isotope \(^{48}{\rm{Ca}}\) has a half-life of about \(2 \times {10^{16}}{\rm{y}}\). (a) A small sample of this isotope is labeled as having an activity of \(1.0\)Ci. What is the mass of the \(^{48}{\rm{Ca}}\)in the sample?
(b) What is unreasonable about this result?
(c) What assumption is responsible?
(a) Calculate the energy released in the\({\rm{\alpha }}\)decay of\({}^{{\rm{238}}}{\rm{U}}\). (b) What fraction of the mass of a single\({}^{{\rm{238}}}{\rm{U}}\)is destroyed in the decay? The mass of\({}^{{\rm{234}}}{\rm{Th}}\)is\(234.043593\,{\rm{u}}\). (c) Although the fractional mass loss is large for a single nucleus, it is difficult to observe for an entire macroscopic sample of uranium. Why is this?
What do you think about this solution?
We value your feedback to improve our textbook solutions.
Explanations 
Textbooks 
Flashcards 
91影视 AI 
Notes 
Study Plans 
Study Sets 
Find a degree 
Magazine