1. ## The half life of a radioactive substance.

I understand part a) as the equation is worked so that it is in terms of k. The solution for part b) is losing me at:

$\lambda=-\frac{ln(2)}{k}*y_0e^k^(^t^1^+ ^\lambda)$

I see that where t was before there is now the symbol lambda but the part after the * symbol is lost on me right now. Why is the equation in terms of k now being multiplied by:

$y_0e^k^(^t^1^+ ^\lambda)$

Thanks for any responses...

2. ## Re: The half life of a radioactive substance.

In part b), it is supposed to be a period, not a multiplication. They are using $\lambda$ as the value of $t$ you found in part a). So, if $y_1 = y_0 e^{kt_1}$, then $y_0 e^{k(t_1+\lambda)} = y_0e^{kt_1}e^{k\lambda}$. Now, replace $y_0e^{kt_1}$ with $y_1$ since they are equal by assumption. So, $y_0e^{kt_1}e^{k\lambda} = y_1e^{k\lambda}$. Next, substitute in the value set for $\lambda$: $y_1e^{k\lambda} = y_1e^{-\ln 2} = \dfrac{1}{2}y_1$.

3. ## Re: The half life of a radioactive substance.

Hey sepoto.

Remember that you are looking at the time t1 + lambda and not t1 (and you also have the information about the half life at t1 and t1 + lambda from the question). Just plug in the information for the lambda + t1 and everything should be ok.

4. ## Re: The half life of a radioactive substance.

So:

$t=\frac{-ln(2)}{k}$

So t might be some number 1590 in years. So if lambda is 1590 then why this equation below. So then "t_1+\lambda" is some time t plus lambda where lambda is the half life of 1590?

$y_0e^{k(t_1+\lambda)}$

5. ## Re: The half life of a radioactive substance.

I'm not exactly sure what you are trying to say: can you please ask again but in more detail (and clarity)?