A radioactive sample has a half life of `40` seconds. When its activity is measured `80` seconds after the beginning, it is found to be `6.932xx10^(18)` dps. During this time total energy released is `6xx10^(8)` joule `(ln2=0.6932)`:-

Options

A. The initial number of atoms in the simple is `1.6xx10^(20)`
B. The initial number of atoms in the sample is `1.6xx10^(21)`
C. Energy released per fission is `5xx10^(-13) J`
D. Energy released per fission is `(5)/(3)xx10^(-13)J`

Explanation

To solve the problem step by step, we will follow the reasoning provided in the video transcript. The question involves a radioactive sample with a known half-life, activity after a certain time, and total energy released. We need to find the energy released per decay (fission).

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Half-life (\(T_{1/2}\)) = 40 seconds
- Activity after 80 seconds (\(A\)) = \(6.932 \times 10^{18}\) dps
- Total energy released = \(6 \times 10^8\) joules
- \(\ln 2 = 0.6932\)

2. **Calculate the Decay Constant (\(\lambda\)):**
The decay constant \(\lambda\) can be calculated using the formula:
\[ \lambda = \frac{\ln 2}{T_{1/2}} = \frac{0.6932}{40 \text{ s}} = 0.01733 \text{ s}^{-1} \]

3. **Use the Activity Formula:**
The activity \(A\) at time \(t\) is given by:
\[ A = A_0 e^{-\lambda t} \]
Rearranging gives:
\[ A_0 = A e^{\lambda t} \]
Substituting the known values:
\[ A_0 = 6.932 \times 10^{18} \times e^{0.01733 \times 80} \]

4. **Calculate \(e^{\lambda t}\):**
First, calculate \(\lambda t\):
\[ \lambda t = 0.01733 \times 80 = 1.3864 \]
Now calculate \(e^{1.3864}\):
\[ e^{1.3864} \approx 4 \]
Thus:
\[ A_0 \approx 6.932 \times 10^{18} \times 4 = 27.728 \times 10^{18} \text{ dps} \]

5. **Calculate the Initial Number of Atoms (\(N_0\)):**
The relationship between activity and the number of radioactive atoms is given by:
\[ A_0 = \lambda N_0 \]
Rearranging gives:
\[ N_0 = \frac{A_0}{\lambda} \]
Substituting the values:
\[ N_0 = \frac{27.728 \times 10^{18}}{0.01733} \approx 1.6 \times 10^{21} \]

6. **Calculate the Number of Decayed Atoms:**
After 80 seconds, the remaining atoms are:
\[ N = \frac{N_0}{4} = \frac{1.6 \times 10^{21}}{4} = 0.4 \times 10^{21} \]
Therefore, the number of decayed atoms is:
\[ N_{decayed} = N_0 - N = 1.6 \times 10^{21} - 0.4 \times 10^{21} = 1.2 \times 10^{21} \]

7. **Calculate Energy Released per Decay:**
The total energy released is given as \(6 \times 10^8\) joules. The energy released per decay (\(E_0\)) is:
\[ E_0 = \frac{\text{Total Energy}}{N_{decayed}} = \frac{6 \times 10^8}{1.2 \times 10^{21}} = 5 \times 10^{-13} \text{ joules} \]

### Final Answer:
The energy released per decay is:
\[ E_0 = 5 \times 10^{-13} \text{ joules} \]