`90%` of a radioactive sample is left undecayed after time `t` has elapesed. What percentage of the initial sample will decay in a total time `2 t` ?

Options

A. 0.2
B. 0.19
C. 0.4
D. 0.38

Explanation

To solve the problem, we need to determine the percentage of the initial radioactive sample that decays after a total time of \(2t\), given that \(90\%\) of the sample is left undecayed after time \(t\).

### Step-by-Step Solution:

1. **Understanding the Decay Formula**:
The number of undecayed nuclei at time \(t\) is given by the formula:
\[ N(t) = N_0 e^{-\lambda t} \]
where \(N_0\) is the initial amount of the sample, \(N(t)\) is the amount remaining after time \(t\), and \(\lambda\) is the decay constant.

2. **Setting Up the Equation for Time \(t\)**:
We know that \(90\%\) of the sample is left undecayed after time \(t\). Thus, we can express this as:
\[ N(t) = 0.9 N_0 \]
Substituting into the decay formula gives:
\[ 0.9 N_0 = N_0 e^{-\lambda t} \]

3. **Simplifying the Equation**:
Dividing both sides by \(N_0\) (assuming \(N_0 \neq 0\)):
\[ 0.9 = e^{-\lambda t} \]

4. **Taking the Natural Logarithm**:
Taking the natural logarithm of both sides:
\[ \ln(0.9) = -\lambda t \]
Rearranging gives:
\[ \lambda = -\frac{\ln(0.9)}{t} \]

5. **Finding the Amount Remaining After Time \(2t\)**:
Now we need to find the amount remaining after time \(2t\):
\[ N(2t) = N_0 e^{-\lambda (2t)} = N_0 e^{-2\lambda t} \]
Substituting \(\lambda\) from the previous step:
\[ N(2t) = N_0 e^{-2 \left(-\frac{\ln(0.9)}{t}\right) t} = N_0 e^{2 \ln(0.9)} = N_0 (0.9)^2 \]

6. **Calculating the Remaining Percentage**:
Now, we calculate \((0.9)^2\):
\[ (0.9)^2 = 0.81 \]
This means that \(81\%\) of the initial sample remains after time \(2t\).

7. **Finding the Decayed Percentage**:
The percentage of the initial sample that has decayed is:
\[ \text{Decayed Percentage} = 100\% - 81\% = 19\% \]

### Final Answer:
Thus, after a total time of \(2t\), \(19\%\) of the initial sample will decay.

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