Calculating half life (Decay Constant)

3 replies · 4K views · last reply
franwilder
Messages
1
Reaction score
0

Homework Statement



The activity of a radioactive nuclide drops 78.5% of it's initial value in 2000 years.


Homework Equations



I know I need to use the decay constant to work out my answer so I have re-arranged the equation as illustrated in my picture, the only problem is I cannot work out the answer and am unsure of which units to use and where they go

The Attempt at a Solution

 
Attachments
  • Handwritten rearrangement of the radioactive decay equation: t = ln(N/N0) divided by -λ, with a circle around it, beside definitions N = number of atoms, N0 = original number of atoms, λ = decay constant, t = time.
    IMG_0606[1].jpg
    35.2 KB · Views: 644
No, you don't need a "decay constant", you just need the definition of "half life". If T is the half life, in years, then after t years the amount left will be [itex]A(1/2)^{t/T}= A2^{-t/T}[/itex] where A is the initial amount at time t= 0.

"The activity of a radioactive nuclide drops 78.5% of it's initial value in 2000 years."

So [itex]A2^{-2000/T}= .785A[/itex]. Can you solve that for T? You will need to take a logarithm of both sides.
 
If the value drops 78.5% then there will only be 21.5% left (just a tad over 2 half lives).
 
franwilder said:

Homework Statement



The activity of a radioactive nuclide drops 78.5% of it's initial value in 2000 years.


Homework Equations



I know I need to use the decay constant to work out my answer so I have re-arranged the equation as illustrated in my picture, the only problem is I cannot work out the answer and am unsure of which units to use and where they go

The Attempt at a Solution


The units of the decay constant are reciprocal years, since the ln of the ratio of the N's is dimensionless. So you first solve for lambda, and then use your equation again to calculate the value of t that makes the ratio of N's equal to 1/2. This is the half life.