Q6 b) i) (1 Viewer)

[Synergie]

Was there enough information to answer that question?

 

[possessionless]

is this the growth and decay question?

 

[Synergie]

Yes, finding the value of k, in the question relating to Radium decay

 

[Thecorey0]

I got k=ln2/1600

 

[AnandDNA]

yeh 0.5=e^-1600k

 

[Ostentatious]

Was there enough information to answer that question?

yep

 

[boxhunter91]

I got k=ln2/1600

Yup thats correct got that aswell.

 

[Joustah]

Subbing in what you know, and knowing the ratio:

Ae^(k x 0) = 2Ae^(k x 1600)
simplify: e^(1600k) = 0.5
solve for k

 

[ShowStopper]

Yup thats correct got that aswell.

yeaah got that too, and got from memory (ii) t= 2000 something years cant remember lol

 

[nosh22]

Yeah I got that too!

 

[raniaaa]

i left out part ii) how did you guys do it

 

[Thecorey0]

i left out part ii) how did you guys do it

Let Q = 1/3

 

[Cataclysm]

Q = Ae^-kt

Q = A/3

.:. A/3 = Ae^-kt

1/3 = e^-kt

loge(1/3) = loge(e)^-kt

loge(1/3) = -ktloge(e)

loge(1/3) = -kt

-[loge(1/3)]/k = t

Which was like... Eh, I forget...

 

[mitchwong650]

Yeah, I think Cataclysm's got it going on,

ends up being: 2536 to the nearest year

 

[Aplus]

Let Q = 1/3

I let Q = 1 and A = 3...

 

[Fortify]

I let Q = 1 and A = 3...

Does the same job