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. `("ln "0.25)/("ln "0.9)h`
B. `("ln "0.5)/("ln "0.81)h`
C. `("ln "0.25)/("ln "0.81)h`
D. None of these

Explanation

To solve the problem of radioactive decay, we will follow these steps:

### Step 1: Set up the differential equation
The decay of the radioactive material is proportional to the amount present. This can be expressed mathematically as:
\[ \frac{dm}{dt} = -km \]
where \( m \) is the mass of the material, \( k \) is the decay constant, and \( t \) is time.

### Step 2: Integrate the equation
To solve the differential equation, we can separate variables:
\[ \frac{dm}{m} = -k \, dt \]
Integrating both sides gives:
\[ \ln |m| = -kt + C \]
where \( C \) is the integration constant.

### Step 3: Solve for the mass function
Exponentiating both sides, we get:
\[ m = e^{-kt + C} = e^C e^{-kt} \]
Let \( m_0 = e^C \), which represents the initial mass. Thus, we can rewrite the equation as:
\[ m(t) = m_0 e^{-kt} \]

### Step 4: Use initial conditions to find \( m_0 \)
Given that the initial mass \( m(0) = 50 \) kg, we have:
\[ m_0 = 50 \]
So, the mass function becomes:
\[ m(t) = 50 e^{-kt} \]

### Step 5: Use the condition after 2 hours to find \( k \)
After 2 hours, the material has lost 10% of its mass, meaning 90% remains:
\[ m(2) = 0.9 \times 50 = 45 \text{ kg} \]
Substituting into the mass function:
\[ 45 = 50 e^{-2k} \]
Dividing both sides by 50:
\[ 0.9 = e^{-2k} \]
Taking the natural logarithm:
\[ \ln(0.9) = -2k \implies k = -\frac{\ln(0.9)}{2} \]

### Step 6: Find the mass remaining after 4 hours
Now, we can find the mass after 4 hours:
\[ m(4) = 50 e^{-4k} \]
Substituting \( k \):
\[ m(4) = 50 e^{-4 \left(-\frac{\ln(0.9)}{2}\right)} = 50 e^{2\ln(0.9)} = 50 (0.9^2) = 50 \times 0.81 = 40.5 \text{ kg} \]

### Step 7: Find the time when the mass is half of the initial mass
We want to find \( t \) when \( m(t) = 25 \) kg:
\[ 25 = 50 e^{-kt} \]
Dividing both sides by 50:
\[ 0.5 = e^{-kt} \]
Taking the natural logarithm:
\[ \ln(0.5) = -kt \implies t = -\frac{\ln(0.5)}{k} \]
Substituting \( k \):
\[ t = -\frac{\ln(0.5)}{-\frac{\ln(0.9)}{2}} = \frac{2 \ln(0.5)}{\ln(0.9)} \]

### Summary of Results
1. The expression for the mass remaining at any time \( t \) is:
\[ m(t) = 50 e^{-kt} \]
2. The mass of the material after 4 hours is:
\[ m(4) = 40.5 \text{ kg} \]
3. The time at which the material has decayed to half of its initial mass is:
\[ t = \frac{2 \ln(0.5)}{\ln(0.9)} \]