Find the decay constant of `.^(55)Co` radio nuclide if its activity is known to decrease `4%` per hour. The decay product is non-radioactive.

Options

A. `1.1xx10^(-5)s^(-1)`
B. `1.10xx10^(-2)s^(-1)`
C. `1.1xx10^(-2)s^(-1)`
D. `1.09xx10^(-5)s^(-1)`

Answer

To find the decay constant (λ) of the radio nuclide \(^{55}\text{Co}\), given that its activity decreases by 4% per hour, we can follow these steps:

### Step 1: Understand the decay process
The activity \(A\) of a radioactive substance is defined as the rate at which it decays. If the activity decreases by 4% per hour, it means that after 1 hour, 96% of the original activity remains.

### Step 2: Express the remaining activity mathematically
If the initial activity is \(A_0\), after 1 hour, the remaining activity \(A\) can be expressed as:
\[ A = A_0 \times (1 - 0.04) = A_0 \times 0.96 \]

### Step 3: Use the exponential decay formula
The activity of a radioactive substance can also be expressed using the decay constant:
\[ A = A_0 e^{-\lambda t} \]
where \(t\) is the time in hours.

### Step 4: Set the equations equal
From the two expressions for activity, we can set them equal to each other:
\[ A_0 \times 0.96 = A_0 e^{-\lambda \cdot 1} \]
We can cancel \(A_0\) from both sides (assuming \(A_0 \neq 0\)):
\[ 0.96 = e^{-\lambda} \]

### Step 5: Take the natural logarithm of both sides
To solve for \(\lambda\), take the natural logarithm:
\[ \ln(0.96) = -\lambda \]
Thus,
\[ \lambda = -\ln(0.96) \]

### Step 6: Calculate \(\lambda\)
Using a calculator, we find:
\[ \ln(0.96) \approx -0.04082 \]
Therefore,
\[ \lambda \approx 0.04082 \text{ hour}^{-1} \]

### Step 7: Convert \(\lambda\) to per second
To convert \(\lambda\) from per hour to per second, we use the conversion factor \(1 \text{ hour} = 3600 \text{ seconds}\):
\[ \lambda \approx \frac{0.04082}{3600} \text{ second}^{-1} \approx 1.1 \times 10^{-5} \text{ second}^{-1} \]

### Final Answer
The decay constant \(\lambda\) of the \(^{55}\text{Co}\) radio nuclide is approximately:
\[ \lambda \approx 1.1 \times 10^{-5} \text{ second}^{-1} \]

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