Question #cf178

2 Answers
May 5, 2017

Answer:

It will never lose ALL of it.

Explanation:

After 1590 years it will have lost 50%
After 2x1590 years only 25% will be left
After 3x1590 years 12.5% will be left.
Etc.

Even after 10x1590=15900 years about 0.1% will be left.

The activity can be represented in a graph (NOT to scale):
graph{10*0.7^x [-4.02, 24.47, -4.12, 10.12]}

May 18, 2017

Answer:

I got a finite value of #t# in years.

Explanation:

If we begin with #A_0# number of atoms of a sample of radioactive material, number of atoms remaining after time #t# is given by the expression

#A(t)=A_0e^(-0.693t/T_(1//2))#
where #T_(1//2)# is half life of the sample

For #color (white).^226"Ra"# whose half life is given the equation reduces to
#A(t)=A_0e^(-0.693t/1590)#
#=>A(t)=A_0e^(-0.000462t)#

Number of atoms in earth are estimated as #=10^50#.
The isotope of radium which is being considered occurs in traces.
If we start with #10^30# number of atoms of radium today, and put the final remaining number as #1#, as last atom of radium can not decay in to its fraction, we get

#1=10^30xxe^(-0.000462t)#
#=>e^(-0.000462t)=10^(-30)#
Taking #ln# of both sides and solving for #t#

#t~~150,000 " "years#
we get a finite value of #t# in years.