Question #2eecb

1 Answer
Alexander L.
May 11, 2016
  1. m(t)=100e^(-ln2/1590t). Mass in grams and time in years.
  2. m(t=1000)=100e^(-ln2/159xx100)
  3. t_(30g)=-ln0.3/ln2xx1590 years
  4. m'(t)=-(10/159ln2)e^(-ln2/1590t)

Explanation:

Nuclear decay is a 1st order reaction .
(dm)/(dt)=-lambdam.
lambda is the decay constant and m is the amount of matter (mass or number of particles).

Solving the differential equation gives m=m_oe^(-lambdat).
m_o is the initial amount of reactant.

To find lambda, the given half life-is 1590 years.

0.5m_o=m_oe^(-lambdaxx1590)
Using ln on both sides of the equation and rearranging it will give lambda=ln2/1590

Answer 1 therefore is m=100e^(-ln2/1590t).
Question 2; just plug in the value t=1000
Question 3; m=30

Question 4; I assume m'(t) is the derivative of m(t) with respect to time;
(dm)/(dt)=d/(dt)100e^(-ln2/1590t)
=-(10/159ln2)e^(-ln2/1590t), t=1000
=-(10/159ln2)e^(-ln2/159xx100)

This gives the rate of change of your radium in 1000 years time.

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