Exponential Integration question

http://i.imgur.com/ilU0H8Q.jpg

For question 15, so far, I have got the following

dN/N = -k dt

From there, I get N(0)e^{-kt}

I solved k to be 0.0001216 that was using k = -ln(1/2)/5700 ==> (ln 2)/5700

then N(t) = N(0)e^{-(ln} ^{2)t/5700}

which is also N(t) = N(0)*-2^{t/5700}

So I am getting close, but what is N(0)?

so to do the part b I solve N(t) = N(0)*-2^{0.4(5700)} - but what is N(0)?

Question 16 I have solved, don't have any issues with that.

Re: Exponential Integration question

There is a little mistake in the indices of the equation you derived, you brought back the minus sign after changing -ln(1/2) to ln(2) so the equation is

$\displaystyle N(0)e^{\frac{ln(2)t}{5700}}$

The 40% is related to number of atoms not time, if it decays to 40% of its original size after a time T then 0.4*N(0)= N(T)

Using the equation you derived,

$\displaystyle 0.4N(0)=N(0)e^{\frac{ln(2)T}{5700}}$

The N(0) cancels on both sides.