Radium decomposes at a rate proportional to the quantity of radium present. It is found that in 25 years, approximately `1*1` precent of a certain quantity of radium has decomposed. Determine approximately how long it will take for one-half of the original amount of radium to decompose ?
`(log_(e)0.989=-0*01106, log_(e)2=0*6931)`

Answer

To solve the problem, we start by modeling the decomposition of radium using a differential equation. The rate of change of the quantity of radium present, \( P(t) \), is proportional to the quantity of radium present itself. We can express this relationship mathematically as:

\[ \frac{dP}{dt} = -kP \]

where \( k \) is a positive constant of proportionality.

### Step 1: Solve the differential equation

We can separate the variables and integrate:

\[ \frac{dP}{P} = -k \, dt \]

Integrating both sides gives:

\[ \ln |P| = -kt + C \]

where \( C \) is the integration constant. Exponentiating both sides, we have:

\[ P = e^{-kt + C} = e^C e^{-kt} \]

Let \( N = e^C \), which represents the initial quantity of radium at \( t = 0 \). Thus, we can rewrite the equation as:

\[ P(t) = N e^{-kt} \]

### Step 2: Use the given information

We know that in 25 years, approximately 1.1% of the radium has decomposed. This means that 98.9% of the original amount remains. Therefore, we can express this mathematically as:

\[ P(25) = 0.989N \]

Substituting into the equation gives:

\[ 0.989N = N e^{-25k} \]

Dividing both sides by \( N \) (assuming \( N \neq 0 \)):

\[ 0.989 = e^{-25k} \]

### Step 3: Solve for \( k \)

Taking the natural logarithm of both sides:

\[ \ln(0.989) = -25k \]

Using the provided value \( \ln(0.989) \approx -0.01106 \):

\[ -0.01106 = -25k \]

Solving for \( k \):

\[ k = \frac{0.01106}{25} \approx 0.0004424 \]

### Step 4: Determine the time for half-life

We want to find the time \( t_{1/2} \) when half of the original amount of radium has decomposed. This means:

\[ P(t_{1/2}) = \frac{N}{2} \]

Substituting into the equation gives:

\[ \frac{N}{2} = N e^{-kt_{1/2}} \]

Dividing both sides by \( N \):

\[ \frac{1}{2} = e^{-kt_{1/2}} \]

Taking the natural logarithm of both sides:

\[ \ln\left(\frac{1}{2}\right) = -kt_{1/2} \]

Using the provided value \( \ln(2) \approx 0.6931 \):

\[ -0.6931 = -kt_{1/2} \]

Substituting \( k \):

\[ -0.6931 = -0.0004424 t_{1/2} \]

Solving for \( t_{1/2} \):

\[ t_{1/2} = \frac{0.6931}{0.0004424} \approx 1567.5 \text{ years} \]

### Final Answer

Thus, it will take approximately **1567.5 years** for one-half of the original amount of radium to decompose.

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