Radioactivity Question, Is this method correct?

[Irishdoug]

Homework Statement

Show using your expression for \lambda that if at a time t1 the amount is y1, then at a time t1 + $\lambda$ the amount will be $\frac{y1}{2}$, no matter what t1 is.

Homework Equations

y = y0 $e^{kt}$

From previous question: half life $\lambda =-ln2/k$

The Attempt at a Solution

y1 = y0 $e^{kt1}$ so t1 = $\frac{lny1}{y0k}$

so:

y (t1+$\lambda$) = y0 e^(k $\frac{lny1}{y0k}$+ k$\frac{-ln2}{k}$) = y0 e^{ln$\frac{y1}{y0}$ +2}

The k's cancel.
-ln2 = ln 1/2
The e and ln cancels

so;
yo * $\frac{y1}{y0}$ * $\frac{1}{2}$ = $\frac{y1}{2}$

 

[Irishdoug]

Fixed the formatting!

 

[gleem]

y (t1+λλ \lambda ) = y0 e^(k lny1y0klny1y0k\frac{lny1}{y0k}+ k−ln2k−ln2k \frac{-ln2}{k}) = y0 e^{lny1y0y1y0\frac{y1}{y0} +2}

This is not correct.

 

[Irishdoug]

I figured it out! Cheers for the response!

 

[Merlin3189]

Homework Statement

Show using your expression for $\lambda$ that if at a time $t_1$ the amount is $y_1$, then at a time $t_1 + \lambda$ the amount will be $\frac{y1}{2}$, no matter what $t_1$ is.

Homework Equations

$$ y_t = y_0 e{(kt)}$$

From previous question: half life $\lambda =\dfrac{-ln2}{k}$

The Attempt at a Solution

$$y_1 = y_0 e^{(kt_1)} \ \ $$

I think the last statement in the first line of 3 is wrong, ie. not $t_1 = \dfrac {ln( y_1)}{y_0 k}$

 

[Irishdoug]

I think the last statement in the first line of 3 is wrong, ie. not $t_1 = \dfrac {ln( y_1)}{y_0 k}$

Hi. What I did instead was:

y0 $e^{kt1+ \lambda}$ --> $\lambda =-ln2/k$

= y0$e^{kt1}$ * $e^{k \lambda}$ =

y1 * $e^{ln(1/2)}$

= $\frac{y1}{2}$