Help!! Rates question (1 Viewer)

[fabl]

I have 10kg of a radioactive material, which appears to continuously decay at a rate of 100% per year. How much will I have after 3 years?


Please explain. The answer is not zero. However, the overall answer is that there will be roughly .49....kg of material left after 3 years. Which confused me as it said that after one year 100% of the 10kg of material decays.


Apparently rates slow down as years go by??

 

[Kaido]

math logic

 

[Kaido]

Jk, it's a simple exponential rate question.
Think logically. Decay does not happen in an instant. It gradually decays, so each time it decays, the factor of decay (i.e. the 100%) changes (more specifically - slows down)
Give it another shot mate, look over your textbook for hints

(havent seen these q's in HSC trials yet)

 

[Rhinoz8142]

Just use the formulae

N= Ae^(kt) BUT MAKE SURE THAT 'k' IS negative because of the DECAY..


its easier than you think

 

[Kaido]

you ruined the fun mate, get out of here Rhinoz

 

[Rhinoz8142]

Why the h8 !

 

[LostInHSC]

Just use the formulae

N= Ae^(kt) BUT MAKE SURE THAT 't' IS negative because of the DECAY..


its easier than you think

Do you mean 'k' is negative? I know it has the same effect but just making sure.

 

[Rhinoz8142]

Do you mean 'k' is negative? I know it has the same effect but just making sure.

oh shit.. yea sorry thats my mistake

 

[InteGrand]

How can you have it decay at 100% per year? Theoretically, it should take an infinite amount of time for it to decay 100%.

 

[Kaido]

How can you have it decay at 100% per year? Theoretically, it should take an infinite amount of time for it to decay 100%.

'theoretically' is not in the yr12 math dictionary im afraid

 

[InteGrand]

Well the original question is ill defined, at any rate (lol).

 

[Kaido]

i think you can actually have it decay at that rate... thus the reason why the answer is non-zero

 

[InteGrand]

Yeah I get what the question's asking now. It's saying the instantaneous rate of decay when the mass remaining is M kg is -M kg/yr .

That is, , where the mass remaining after t years is M kg.

This initial value problem as the solution M(t) = 10e^-t.

So M(3) = 10e^-3 = 0.49787.... ≈ 0.50 (2 d.p.).

 

[Kaido]

Yeah a good trick question trollface.jpg