A certain radioactive material is known to decay at a rate proportional to the amount present. If initially there is 50 kg of the material present and after two hours it is observed that the material has lost 10% of its original mass, find () and expression for the mass of the material remaining at any time t, (ii) the mass of the material after four hours and (iii) the time at which the material has decayed to one half of its initial mass.

Options

A. `N=50e^((1//2)("ln "9)t)`
B. `50e^((1//4)("ln "9)t)`
C. `N=50e^(-("ln "0.9)t)`
D. None of these

Answer

A

Explanation

Let N denote the amount of material present and linear at time t. Then, `(dN)/(dt) =-kN=0`
This differential equation is separable and linear. Its solution is `N=ce^(kt)`
At t=0, we are given that `N=50`. Therefore, from equation (1), `50=ce^(k(0))` or `c=50`.
Thus, `N=50e^(kt)`............(2)
At `t=2, 10%` of the original mass of 50 mg or 5 mg has decayed.
Hence, at `t=2, N=50-5=45`
Sustituting these values into equation (2) and solving for k,
We have `45=50e^(2k)` or `k =1/2 log(45/50)`
Substituting this value into (2), we obtain the amount of mass present at any time t as
`N=50e^((1//2)("ln "0.9)t)`
where t is measured in hours. ...............(3)
where t is measured in hours.