Calculating Half-Life Using Calculus: Deriving λ from y = y0e-kt

[ttttrigg3r]

Homework Statement

This is the equation
y = y₀e^-kt

Homework Equations

b) Show using your expression for λ that if at time t1 the amount is y1, then at time t1 + λ it will be y1/2, no matter what t1 is.

The Attempt at a Solution

y₁=y₀e^kt₁

that part I got. Then the answer key goes into the next step saying:λ = (−ln2/k)y₀e^k(t₁+λ)

That is where I got lost. What is the step to go from the y1 equation into the lambda equation? I know that for half life, y1=y0/2 so that makes t=ln2/k . The time it takes for an element to decompose to half its mass is t=ln2/k . How do I make the connection to the very last lambda equation?

 

[ehild]

The statement "at time t1 the amount is t1" is the same as y1=y₀e^-kt1. At any time, the amount is y=y₀e^-kt.
At time t=t1+λ, the amount is y=y₀e^-k(t1+λ)=y₀e^-kt1-kλ. You know the identity a^x+y=a^x a^y, don't you? And λ is the half-life, so y₀e^-kλ=y₀/2.
Can you proceed from here?


ehild

 

[ttttrigg3r]

using your method and identity, I got: y=y₀e^-kt1*e^-k\lambda
so
y=y₁*e^-k\lambda

and then I am stuck again. lambda = half life, but what does that mean for me to get to y0e-kλ=y0/2.

 

[ehild]

y₀e^-kλ=y₀/2, that is e^-kλ=1/2.

ehild

 

[ttttrigg3r]

I still do not see it. is lambda in this equation being treated as a variable or constant? is lambda the same as time?

 

[SammyS]

λ is the time it takes for a sample of an element to decompose so that the mass of the original element in the resulting sample is half of the original mass. Thus, λ, must be the half-life.

 

[ttttrigg3r]

Ok I think I got it. This is my understanding, tell me if I am correct.

e^-k\lambda is the multiplier it takes to make an element exactly 1/2 of itself therefore we can set e^-k\lambda=1/2
So when I come to y=y₀e^-kt₁*e^-kλ That is the same thing as y=y₀e^-kt₁*(1/2) and knowing that y₀e^-kt₁=y₁ we can say y=y₁*(1/2)

am I right?

 

[ehild]

That is the same thing as y=y₀e^-kt₁*(1/2) and knowing that y₀e^-kt₁=y₁ we can say y=y₁*(1/2)

am I right?

Yes, You are right, well done!

ehild

 

[ttttrigg3r]

THank you all those who helped.