The rate of decay of particular isotope of Radium (in mg per century) is proportional to its mass (in mg). A 50mg sample takes one century to decay to 48mg. Ho0w long will it take before there are 45 mg of the sample?

Let us define the following variables:

# { (x,"mass of Radium (mg)"), (t, "time (centuries)") :} #

Then

# -dx/dt prop x => dx/dt = -kx #

where #k# is the constant of proportionality. This is a First Order separable Differential Equation and we can separate the variables to get:

# int \ 1/x \ dx = int \ -k \ dt #

Which we can integrate to get:

# \ \ \ ln |x| = -kt + C #
# :. ln x = -kt + C # , as #x# is positive

We initially started off with #x=50# (mg) #=>x=50# when #t=0#, so we can substitute into the DE solution to get:

# ln 50 = C #

We are also told that #x=48# (mg) when #t=1# (century) so we can substitute into the DE solution to get:

# ln 48 =-k + ln50 => k = ln50-ln48 = ln(50/48) #
# :. k = ln(25/24) #

And so the Specific Solution is:

# ln x = -tln(25/24) + ln 50 # ....[1]
# :. ln x = ln(25/24)^(-t) + ln 50 #
# :. ln x = ln (50(25/24)^(-t)) #
# :. x = 50(25/24)^(-t) #
# :. x = 50(24/25)^t #

[ We should just check that we have not made a mistake by checking the initial condition:

#t=0 => x=50(24/25)^0=50#
#t=1 => x=50(24/25)^1=24#

so we know the solution is sound]

We are asked to find #t# when #x=45#, and so using [1] we have:

# ln 45 = -tln(25/24) + ln 50 #
# :. tln(25/24) = ln 50 - ln45 #
# :. tln(25/24) = ln 50/45 #
# :. tln(25/24) = ln 10/9 #
# :. t = (ln 10/9)/(ln(25/24)) #
# :. t = 2.58097 ... #

Hence it will take 2.6 centuries.