Radioactive Decay: Dependency on Temp & Pressure?

AI Thread Summary
Radioactive decay is primarily a property of the atomic nucleus and is not significantly influenced by temperature or pressure. While temperature can increase particle movement, which may affect reaction rates in some contexts, it does not alter the fundamental decay process. Pressure does not play a role in radioactive decay according to current reactivity equations. The discussion highlights the confusion surrounding how these factors interact with atomic behavior. Overall, the consensus is that radioactive decay remains largely independent of external temperature and pressure conditions.
soccerguy312
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Is radioactive decay dependant on temperature? what about pressure?

my initial thinking is yes and no
 
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soccerguy312 said:
Is radioactive decay dependant on temperature? what about pressure?

my initial thinking is yes and no

Welcome to the PF. Why do you think those are the answers. What sources of information did you check out in order to get to those answers? What physical mechanisms are you thinking are behind any change versus temperature and pressure?
 
because pressure has no feed into any reactivity equations I have seen. however temperature increases how fast particles move.. and temperature affects pressure so I am confused
 
Radioactive decay is a property/phenomenon of the nucleus.

Temperature and pressure affect the atom, and are related to the translational motion of atoms., or in the case of a compressive stress, pressure relates to a local force applied to a unit area.
 
Thread 'Confusion regarding a chemical kinetics problem'

Thread 'Confusion regarding a chemical kinetics problem'

TL;DR Summary: cannot find out error in solution proposed. [![question with rate laws][1]][1] Now the rate law for the reaction (i.e reaction rate) can be written as: $$ R= k[N_2O_5] $$ my main question is, WHAT is this reaction equal to? what I mean here is, whether $$k[N_2O_5]= -d[N_2O_5]/dt$$ or is it $$k[N_2O_5]= -1/2 \frac{d}{dt} [N_2O_5] $$ ? The latter seems to be more apt, as the reaction rate must be -1/2 (disappearance rate of N2O5), which adheres to the stoichiometry of the...
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