A sample which has half life of `10^(33)` year . If initial number of nuclei of the sample is `26 xx 10^(24)`. Then find out of the number of nuclei decayed in 1 year.

Options

A. `1.82 xx10^(-7)`
B. `182 xx 10^(-7)`
C. `18.2 xx 10^(-7)`
D. `1820 xx 10^(-7)`

Explanation

To solve the problem, we need to find the number of nuclei that decay in one year for a sample with a half-life of \(10^{33}\) years and an initial number of nuclei of \(26 \times 10^{24}\).

### Step-by-Step Solution:

1. **Identify Given Values:**
- Half-life \(T = 10^{33}\) years
- Initial number of nuclei \(N_0 = 26 \times 10^{24}\)
- Time period \(t = 1\) year

2. **Determine the Decay Constant (\(\lambda\)):**
The decay constant \(\lambda\) is related to the half-life by the formula:
\[ \lambda = \frac{\ln(2)}{T} \]
Substituting the value of \(T\):
\[ \lambda = \frac{\ln(2)}{10^{33}} \approx \frac{0.693}{10^{33}} \approx 6.93 \times 10^{-34} \text{ years}^{-1} \]

3. **Use the Decay Formula:**
The number of nuclei decayed in a small time interval \(dt\) is given by:
\[ -dN = \lambda N_0 dt \]
For \(dt = 1\) year, we can substitute the values:
\[ -dN = \lambda N_0 \cdot 1 \]
Thus,
\[ dN = \lambda N_0 \]

4. **Calculate the Number of Decayed Nuclei:**
Substitute \(\lambda\) and \(N_0\):
\[ dN = \left(6.93 \times 10^{-34}\right) \times \left(26 \times 10^{24}\right) \]
\[ dN \approx 6.93 \times 26 \times 10^{-10} = 180.18 \times 10^{-10} \approx 1.8018 \times 10^{-8} \]

5. **Final Result:**
The number of nuclei decayed in 1 year is approximately:
\[ dN \approx 1.8 \times 10^{-8} \]

### Final Answer:
The number of nuclei decayed in 1 year is approximately \(1.8 \times 10^{-8}\).

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