Question

Question #2eecb

Answer 1

1. `m(t)=100e^(-ln2/1590t)`. Mass in grams and time in years.
2. `m(t=1000)=100e^(-ln2/159xx100)`
3. `t_(30g)=-ln0.3/ln2xx1590 years`
4. `m'(t)=-(10/159ln2)e^(-ln2/1590t)`

Explanation:

Nuclear decay is a `1st` order reaction .
`(dm)/(dt)=-lambdam`.
`lambda` is the decay constant and `m` is the amount of matter (mass or number of particles).

Solving the differential equation gives `m=m_oe^(-lambdat)`.
`m_o` is the initial amount of reactant.

To find `lambda`, the given half life-is `1590` years.

`0.5m_o=m_oe^(-lambdaxx1590)`
Using `ln` on both sides of the equation and rearranging it will give `lambda=ln2/1590`

Answer `1` therefore is `m=100e^(-ln2/1590t)`.
Question 2; just plug in the value `t=1000`
Question 3; `m=30`

Question 4; I assume `m'(t)` is the derivative of `m(t)` with respect to time;
`(dm)/(dt)=d/(dt)100e^(-ln2/1590t)`
`=-(10/159ln2)e^(-ln2/1590t)`, `t=1000`
`=-(10/159ln2)e^(-ln2/159xx100)`

This gives the rate of change of your radium in 1000 years time.