Question

The amount A of the radioactive element radium in a sample decays at a rate proportional to the amount of radium present. Given the half-life of radium is 1690 years: (a) Write a differential equation that models the amount A of radium present at time t (b) Find the general solution of the differential equation. (c) Find the particular solution of the differential equation with the initial condition A(0)=10 g. (d) How much radium will be present in the sample at t=300 years?

Answer 1

Answer:

a) -dm/m = k*dt

b) m=m₀*e^(-k*t)

c) m=10 g*e^(-4.1* 10⁻⁴ years⁻¹*t)

d) m= 8.842 gr

( same results obtained previously by Xero099)

Step-by-step explanation:

Since the amount of radium is proportional to its quantity . Denoting as m the mass of Radium , and t as time. we have:

a) -dm/dt = k*m , where k is the proportionality constant

b) solving the differential equation;

-dm/m = k*dt

∫dm/m = -∫k*dt

ln m = -k*t + C

for t=0 , we have m=m₀ (initial mass)

then

ln m₀ = 0 + C → C=ln m₀

therefore

ln (m/m₀) = -k*t

m=m₀*e^(-k*t)

c) for t= 1690 years , the quantity of radium is half , then m=m₀/2 , thus

ln (m₀/2/m₀) = -k*t

ln (1/2) =   -k*t

(ln 2 )/t = k

k =  ln 2  / 1690 years = 4.1* 10⁻⁴ years⁻¹

then for m₀= 10 g

m=10 g*e^(-4.1* 10⁻⁴ years⁻¹*t)

d) at t= 300 years

m=10 g*e^(-4.1* 10⁻⁴ years⁻¹*300 years) = 8.842 gr

then the mass is m= 8.842 gr

Answer 2

Answer:

a) $\frac{dm}{dt} = -k\cdot m$, b) $m(t) = m_{o}\cdot e^{-\frac{t}{\tau} }$, c) $m(t) = 10\cdot e^{-\frac{t}{2438.155} }$, d) $m(300) \approx 8.842\,g$

Step-by-step explanation:

a) Let assume an initial mass m decaying at a constant rate k throughout time, the differential equation is:

$\frac{dm}{dt} = -k\cdot m$

b) The general solution is found after separating variables and integrating each sides:

$m(t) = m_{o}\cdot e^{-\frac{t}{\tau} }$

Where $\tau$ is the time constant and $k = \frac{1}{\tau}$

c) The time constant is:

$\tau = \frac{1690\,yr}{\ln 2}$

$\tau = 2438.155\,yr$

The particular solution of the differential equation is:

$m(t) = 10\cdot e^{-\frac{t}{2438.155} }$

d) The amount of radium after 300 years is:

$m(300) \approx 8.842\,g$