Question #2eecb

1 Answer
May 11, 2016

Answer:

  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.