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?

Question

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?

Isaiah Allen · Accepted Answer

Amount of Radium after #t# centuries is # 50(24/25)^t #

It will take 2.6 centuries for the Radium to weigh 45mg.

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.

William Abdalla · Answer

undefined We can model the rate of decay using the differential equation:

dm/dt = -k * m

where m is the mass of the isotope at time t, and k is the decay constant.

Given that the rate of decay is proportional to the mass, we can write:

dm/dt = km

Using the initial condition that a 50 mg sample decays to 48 mg in one century:

dm/dt = k * 50, when m = 50

So, k = (dm/dt) / 50 = (48 - 50) / 100 = -0.02 mg per century.

The solution to the differential equation is:

m(t) = m(0) * e^(-kt)

We know that when t = 0, m(0) = 50 mg, so:

m(t) = 50 * e^(-0.02t)

Now, we want to find how long it takes for the sample to decay to 45 mg:

45 = 50 * e^(-0.02t)

Divide both sides by 50:

0.9 = e^(-0.02t)

Taking the natural logarithm of both sides:

ln(0.9) = -0.02t

Solve for t:

t = ln(0.9) / -0.02 ≈ 17.33 centuries

So, it will take approximately 17.33 centuries before there are 45 mg of the sample.