A radioactive sample of half life 10 days contains 1000X nuclei. Number of original nuclei present after 5 days is

Options

A. 707 x
B. 750 x
C. 500 x
D. 250 x

Answer

To find the number of original nuclei present after 5 days in a radioactive sample with a half-life of 10 days, we can use the formula for radioactive decay:

\[ A = A_0 \left( \frac{1}{2} \right)^{\frac{t}{t_{1/2}}} \]

Where:
- \( A \) is the number of active nuclei after time \( t \).
- \( A_0 \) is the initial number of nuclei.
- \( t \) is the elapsed time.
- \( t_{1/2} \) is the half-life of the substance.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Initial number of nuclei, \( A_0 = 1000x \)
- Half-life, \( t_{1/2} = 10 \) days
- Time elapsed, \( t = 5 \) days

2. **Substitute the Values into the Formula:**
\[ A = 1000x \left( \frac{1}{2} \right)^{\frac{5}{10}} \]

3. **Simplify the Exponent:**
\[ \frac{5}{10} = \frac{1}{2} \]
So, the equation becomes:
\[ A = 1000x \left( \frac{1}{2} \right)^{\frac{1}{2}} \]

4. **Calculate \( \left( \frac{1}{2} \right)^{\frac{1}{2}} \):**
\[ \left( \frac{1}{2} \right)^{\frac{1}{2}} = \frac{1}{\sqrt{2}} \approx 0.707 \]

5. **Substitute Back to Find \( A \):**
\[ A = 1000x \times 0.707 \]

6. **Final Calculation:**
\[ A \approx 707x \]

### Conclusion:
The number of original nuclei present after 5 days is approximately \( 707x \).