The element curium `_96^248 Cm` has a mean life of `10^13s`. Its primary decay modes are spontaneous fission and `alpha`-decay, the former with a probability of 8% and the later with a probability of 92%, each fission releases 200 MeV of energy. The masses involved in decay are as follows
`_96^248 Cm=248.072220 u`,
`_94^244 P_u=244.064100 u` and `_2^4 He=4.002603u`. Calculate the power output from a sample of `10^20` Cm atoms. (`1u=931 MeV//c^2`)

Explanation

To solve the problem, we need to calculate the power output from a sample of \(10^{20}\) atoms of curium \(_{96}^{248}Cm\). We will follow these steps:

### Step 1: Calculate the total energy released from decay

The total energy released from decay can be calculated by considering both decay modes: spontaneous fission and alpha decay.

1. **Energy from spontaneous fission**:
- Probability of fission = 8% = 0.08
- Energy released per fission = 200 MeV

The number of fissions from \(10^{20}\) atoms:
\[ N_f = 0.08 \times 10^{20} = 8 \times 10^{18} \]

Total energy from fission:
\[ E_f = N_f \times \text{Energy per fission} = 8 \times 10^{18} \times 200 \text{ MeV} = 1.6 \times 10^{21} \text{ MeV} \]

2. **Energy from alpha decay**:
- Probability of alpha decay = 92% = 0.92
- We need to calculate the energy released in alpha decay using the mass defect.

The mass defect for alpha decay can be calculated as follows:
\[ \text{Mass defect} = m_{Cm} - (m_{Pu} + m_{He}) = 248.072220 \, u - (244.064100 \, u + 4.002603 \, u) = 0.005517 \, u \]

Converting mass defect to energy:
\[ E_{\alpha} = \text{Mass defect} \times 931 \text{ MeV/u} = 0.005517 \, u \times 931 \text{ MeV/u} \approx 5.14 \text{ MeV} \]

The number of alpha decays from \(10^{20}\) atoms:
\[ N_{\alpha} = 0.92 \times 10^{20} = 9.2 \times 10^{19} \]

Total energy from alpha decay:
\[ E_{\alpha} = N_{\alpha} \times E_{\alpha} = 9.2 \times 10^{19} \times 5.14 \text{ MeV} \approx 4.73 \times 10^{20} \text{ MeV} \]

3. **Total energy released**:
\[ E_{\text{total}} = E_f + E_{\alpha} = 1.6 \times 10^{21} \text{ MeV} + 4.73 \times 10^{20} \text{ MeV} \approx 2.07 \times 10^{21} \text{ MeV} \]

### Step 2: Convert total energy to joules

To convert MeV to joules, we use the conversion factor \(1 \text{ MeV} = 1.6 \times 10^{-13} \text{ J}\):
\[ E_{\text{total}} \text{ (in Joules)} = 2.07 \times 10^{21} \text{ MeV} \times 1.6 \times 10^{-13} \text{ J/MeV} \approx 3.312 \times 10^{8} \text{ J} \]

### Step 3: Calculate the power output

Power is defined as energy per unit time. Given the mean life of curium is \(10^{13} \text{ s}\):
\[ P = \frac{E_{\text{total}}}{\text{mean life}} = \frac{3.312 \times 10^{8} \text{ J}}{10^{13} \text{ s}} \approx 3.312 \times 10^{-5} \text{ W} \]

### Final Answer
The power output from a sample of \(10^{20}\) curium atoms is approximately:
\[ P \approx 3.31 \times 10^{-5} \text{ W} \]