A 280 day old radioactive substances shows an activity of 6000 dps, 140 days later its activity becomes 3000 dps. What was its initial activity ?

Options

A. 20000 dps
B. 24000 dps
C. 12000 dps
D. 6000 dps

Answer

To find the initial activity of the radioactive substance, we can use the exponential decay formula for radioactive decay, which is given by:

\[ A(t) = A_0 e^{-\lambda t} \]

where:
- \( A(t) \) is the activity at time \( t \),
- \( A_0 \) is the initial activity,
- \( \lambda \) is the decay constant,
- \( t \) is the time elapsed.

### Step 1: Set up the equations for the two time points

1. At \( t = 280 \) days, the activity \( A(280) = 6000 \) dps:
\[ 6000 = A_0 e^{-\lambda \cdot 280} \]

2. At \( t = 420 \) days (which is 140 days later), the activity \( A(420) = 3000 \) dps:
\[ 3000 = A_0 e^{-\lambda \cdot 420} \]

### Step 2: Divide the two equations

To eliminate \( A_0 \), we can divide the first equation by the second equation:

\[ \frac{6000}{3000} = \frac{A_0 e^{-\lambda \cdot 280}}{A_0 e^{-\lambda \cdot 420}} \]

This simplifies to:

\[ 2 = e^{-\lambda \cdot 280} \cdot e^{\lambda \cdot 420} \]

Using the property of exponents, we can rewrite this as:

\[ 2 = e^{\lambda \cdot (420 - 280)} = e^{\lambda \cdot 140} \]

### Step 3: Take the natural logarithm of both sides

Taking the natural logarithm of both sides gives:

\[ \ln(2) = \lambda \cdot 140 \]

From this, we can solve for \( \lambda \):

\[ \lambda = \frac{\ln(2)}{140} \]

### Step 4: Substitute \( \lambda \) back into one of the original equations

Now, we can substitute \( \lambda \) back into one of the original equations to find \( A_0 \). We will use the second equation:

\[ 3000 = A_0 e^{-\lambda \cdot 420} \]

Substituting \( \lambda \):

\[ 3000 = A_0 e^{-\left(\frac{\ln(2)}{140}\right) \cdot 420} \]

This simplifies to:

\[ 3000 = A_0 e^{-3 \ln(2)} = A_0 \cdot e^{\ln(2^{-3})} = A_0 \cdot \frac{1}{8} \]

### Step 5: Solve for \( A_0 \)

Now, we can solve for \( A_0 \):

\[ A_0 = 3000 \cdot 8 = 24000 \text{ dps} \]

### Conclusion

The initial activity \( A_0 \) of the radioactive substance was **24000 dps**.

---