A nucleus of `Ux_1` has a half life of 24.1 days. How long a sample of `Ux_1` will take to change to 90% of `Ux_2`.

Options

A. 80 days
B. 40 days
C. 20 days
D. 10 days

Explanation

To solve the problem, we need to determine how long it will take for a sample of `Ux_1` to decay to 90% of `Ux_2`. Here are the steps to arrive at the solution:

### Step 1: Calculate the decay constant (λ)
The decay constant (λ) can be calculated using the formula:
\[ \lambda = \frac{0.693}{T_{1/2}} \]
where \(T_{1/2}\) is the half-life of the substance. Given that the half-life of `Ux_1` is 24.1 days, we can substitute this value into the formula:
\[ \lambda = \frac{0.693}{24.1 \text{ days}} \approx 0.0287 \text{ per day} \]

### Step 2: Determine the remaining amount of `Ux_1`
If 90% of the sample has changed to `Ux_2`, then only 10% of the original sample of `Ux_1` remains. If we denote the initial amount of `Ux_1` as \(N_0\), then after some time \(t\), the amount remaining \(N\) is:
\[ N = N_0 - 0.9N_0 = 0.1N_0 \]

### Step 3: Use the exponential decay formula
The amount of a radioactive substance remaining after time \(t\) can be expressed using the exponential decay formula:
\[ N = N_0 e^{-\lambda t} \]
Substituting the expression for \(N\) from Step 2, we get:
\[ 0.1N_0 = N_0 e^{-\lambda t} \]

### Step 4: Simplify the equation
We can cancel \(N_0\) from both sides (assuming \(N_0 \neq 0\)):
\[ 0.1 = e^{-\lambda t} \]

### Step 5: Take the natural logarithm of both sides
Taking the natural logarithm (ln) of both sides gives:
\[ \ln(0.1) = -\lambda t \]

### Step 6: Solve for time \(t\)
Rearranging the equation to solve for \(t\):
\[ t = -\frac{\ln(0.1)}{\lambda} \]
Substituting the value of \(\lambda\):
\[ t = -\frac{\ln(0.1)}{0.0287} \]

### Step 7: Calculate \(t\)
Using the fact that \(\ln(0.1) \approx -2.302\):
\[ t = -\frac{-2.302}{0.0287} \approx 80.2 \text{ days} \]

Thus, it will take approximately **80 days** for the sample of `Ux_1` to change to 90% of `Ux_2`.