Number of nuclei of a radioactive substance are `1000` and `900` at times `t=0` and time `t=2 s`. Then, number of nuclei at time `t=4s` will be

Options

A. 8000
B. 810
C. 790
D. 700

Answer

To solve the problem of finding the number of nuclei of a radioactive substance at time \( t = 4 \) seconds, we will use the radioactive decay equation:

\[ N(t) = N_0 e^{-\lambda t} \]

where:
- \( N(t) \) is the number of nuclei at time \( t \),
- \( N_0 \) is the initial number of nuclei,
- \( \lambda \) is the decay constant,
- \( t \) is the time.

### Step 1: Identify the initial conditions
At \( t = 0 \), the number of nuclei \( N_0 = 1000 \).
At \( t = 2 \) seconds, the number of nuclei \( N(2) = 900 \).

### Step 2: Set up the equation for \( t = 2 \) seconds
Using the decay equation for \( t = 2 \):

\[ 900 = 1000 e^{-\lambda \cdot 2} \]

### Step 3: Solve for \( e^{-\lambda \cdot 2} \)
Divide both sides by 1000:

\[ \frac{900}{1000} = e^{-2\lambda} \]

This simplifies to:

\[ 0.9 = e^{-2\lambda} \]

### Step 4: Take the natural logarithm of both sides
Taking the natural logarithm:

\[ \ln(0.9) = -2\lambda \]

### Step 5: Solve for \( \lambda \)
Rearranging gives:

\[ \lambda = -\frac{1}{2} \ln(0.9) \]

### Step 6: Set up the equation for \( t = 4 \) seconds
Now we want to find \( N(4) \):

\[ N(4) = 1000 e^{-\lambda \cdot 4} \]

### Step 7: Substitute \( \lambda \) into the equation
Substituting \( \lambda \):

\[ N(4) = 1000 e^{-4 \left(-\frac{1}{2} \ln(0.9)\right)} \]

This simplifies to:

\[ N(4) = 1000 e^{2 \ln(0.9)} \]

### Step 8: Simplify the expression
Using the property of exponents:

\[ N(4) = 1000 (0.9)^2 \]

### Step 9: Calculate \( (0.9)^2 \)
Calculating \( (0.9)^2 \):

\[ (0.9)^2 = 0.81 \]

### Step 10: Calculate \( N(4) \)
Now substitute back:

\[ N(4) = 1000 \times 0.81 = 810 \]

Thus, the number of nuclei at \( t = 4 \) seconds is:

\[ \boxed{810} \]