A radioactive sample remains undecayed `9/16` after time t.How much sample remains undecayed after time `t/2`

Options

A. `3/4`
B. `9/16`
C. `4/3`
D. `16/9`

Answer

To solve the problem, we need to use the concept of radioactive decay. The number of undecayed nuclei at any time \( t \) can be expressed using the exponential decay formula:

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

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

### Step-by-Step Solution

1. **Understand the given information**: We know that after time \( t \), the sample remains undecayed \( \frac{9}{16} \) of its initial amount. This can be expressed as:
\[ N(t) = \frac{9}{16} N_0 \]

2. **Set up the equation using the decay formula**: From the decay formula, we can write:
\[ N(t) = N_0 e^{-\lambda t} \]
Therefore, we can equate:
\[ \frac{9}{16} N_0 = N_0 e^{-\lambda t} \]

3. **Cancel \( N_0 \)**: Assuming \( N_0 \neq 0 \), we can divide both sides by \( N_0 \):
\[ \frac{9}{16} = e^{-\lambda t} \]

4. **Take the natural logarithm**: To solve for \( \lambda t \), take the natural logarithm of both sides:
\[ -\lambda t = \ln\left(\frac{9}{16}\right) \]
Thus,
\[ \lambda t = -\ln\left(\frac{9}{16}\right) \]

5. **Find the undecayed sample at time \( \frac{t}{2} \)**: We need to find the amount of undecayed sample at time \( \frac{t}{2} \):
\[ N\left(\frac{t}{2}\right) = N_0 e^{-\lambda \frac{t}{2}} \]

6. **Substituting \( \lambda \frac{t}{2} \)**: From the earlier step, we know:
\[ \lambda \frac{t}{2} = \frac{1}{2} \left(-\ln\left(\frac{9}{16}\right)\right) = -\frac{1}{2} \ln\left(\frac{9}{16}\right) \]
Therefore,
\[ N\left(\frac{t}{2}\right) = N_0 e^{\frac{1}{2} \ln\left(\frac{9}{16}\right)} \]

7. **Simplifying the expression**: We can simplify this using properties of exponents:
\[ N\left(\frac{t}{2}\right) = N_0 \left(\frac{9}{16}\right)^{\frac{1}{2}} = N_0 \cdot \frac{3}{4} \]

8. **Final answer**: The amount of undecayed sample after time \( \frac{t}{2} \) is:
\[ N\left(\frac{t}{2}\right) = \frac{3}{4} N_0 \]

### Conclusion
Thus, after time \( \frac{t}{2} \), the sample remains undecayed \( \frac{3}{4} \) of its initial amount.