A sample of radioactive material contains 1.00e^15 atoms and has an activity of 6.00e^11 Bq. What is the half life?

Question

anonymous · Answer

$$A=A_0 e^{rt}$$

$$.5= e^{(-6.00 e^{11})t}$$

natural log both side and solve for t
$$1.92946*10^{-6}$$

I would take it with grain of salt,

anonymous · Answer

i have 1.16e^3 s..may be you have minutes.let me convert

anonymous · Answer

no i mean the answer in the book(sorry)

anonymous · Answer

i think it is e^(negative power)

anonymous · Answer

but yh u have -ive already

anonymous · Answer

i meant r= -r

anonymous · Answer

can you recheck all the numbers

anonymous · Answer

ok lemme see

anonymous · Answer

I have the same formula but instead of r the book uses lambda. and it is negative

anonymous · Answer

so did the book used $$\left(6* 10^{-4}\right)$$for r?

anonymous · Answer

Ok so what i have here is R=Ro e^-(lambda)t

anonymous · Answer

What did they plug in for lambda

anonymous · Answer

lambda= Ro/No

anonymous · Answer

No= $$5.98741\times 10^8$$

lambda=6* 10^-4
$$\frac{1}{2}\text{==}E^{-\left(6\ 10^{-4}\right)t}$$

anonymous · Answer

wow nice!!

anonymous · Answer

thanks a lot:)

anonymous · Answer

not really sure what N0 means , but it works

anonymous · Answer

No= number of undecayed atoms

anonymous · Answer

now back to my own physics