The element curium 248_96Cm has a mean life of 10^13 s. Its primary decay modes are spontaneous fiss

Q: Why do we use mean life instead of half-life to find activity?

Activity is defined as $A = \lambda N$. The decay constant $\lambda$ is exactly the reciprocal of the mean life ($1/t_{\text{mean}}$). If half-life was given, we would use $\lambda = \ln(2)/t_{1/2}$.

Q: Why do we multiply the total activity by the probabilities?

The total activity gives the total number of decays per second. Since 8% of these are fission and 92% are alpha decays, multiplying by 0.08 and 0.92 gives the specific number of each type of decay occurring every second.

Q: How did we get the energy of the alpha decay?

We used Einstein's mass-energy equivalence principle. The mass defect (difference in mass before and after decay) is converted into energy. Multiplying the mass defect in atomic mass units (u) by 931 gives the energy in MeV.

Q: Why is the energy from fission so much higher than alpha decay?

Spontaneous fission involves the nucleus splitting into two roughly equal halves, releasing a massive amount of binding energy. Alpha decay only emits a small helium nucleus, resulting in a much smaller energy release.

Q: How do we convert MeV/s to Watts?

1 Watt is equal to 1 Joule per second. To convert MeV to Joules, we multiply by $1.6 \times 10^{-13}$ (since $1 \text{ eV} = 1.6 \times 10^{-19} \text{ J}$ and $1 \text{ MeV} = 10^6 \text{ eV}$).

Question

The element curium $^{248}_{96}	ext{Cm}$ has a mean life of $10^{13}	ext{ s}$. Its primary decay modes are spontaneous fission and $\alpha$-decay, the former with a probability of 8% and the latter with a probability of 92%, each fission releases 200 MeV of energy. The masses involved in decay are as follows :
$^{248}_{96}	ext{Cm} = 248.072220	ext{ u}$,
$^{244}_{94}	ext{Pu} = 244.064100	ext{ u}$ and $^{4}_{2}	ext{He} = 4.002603	ext{ u}$. Calculate the power output from a sample of $10^{20}	ext{ Cm}$ atoms.
($1	ext{ u} = 931	ext{ MeV}/c^2$)

Sigma Percentile · Accepted Answer

### The Dual Nature of Curium's Decay

Imagine you are holding a sample of Curium-248, containing a staggering $10^{20}$ atoms. This sample is not sitting idle; it is actively decaying. Interestingly, it doesn't just decay in one way. It has a dual personality: 92% of the time, it undergoes $\alpha$-decay, gently spitting out a helium nucleus. But 8% of the time, it undergoes spontaneous fission, violently splitting into two large fragments and releasing a massive $200	ext{ MeV}$ of energy per event.

Our goal is to find the total power output of this sample. Power is simply the total energy released per second. To find this, we need to know two things: how many decays are happening every second, and how much energy each type of decay releases.

### Calculating the Activity

First, let's figure out the total rate of decay, also known as the Activity ($A$). The activity is given by the formula $A = \lambda N$, where $\lambda$ is the decay constant and $N$ is the number of atoms.

We are given the mean life ($t_{	ext{mean}}$) of the sample, which is $10^{13}	ext{ s}$. The decay constant is simply the reciprocal of the mean life, so $\lambda = \frac{1}{t_{	ext{mean}}}$.

Substituting this into our activity formula gives:

$$A = \frac{N}{t_{	ext{mean}}}$$

Plugging in our values:

$$A = \frac{10^{20}}{10^{13}} = 10^7 	ext{ disintegrations/s}$$

This means that every single second, ten million Curium atoms are decaying!

### The Energy of Alpha Decay

We already know that each fission event releases $200	ext{ MeV}$. But what about the $\alpha$-decay? To find this, we need to calculate the mass defect ($\Delta m$) of the reaction:

$$^{248}_{96}	ext{Cm} ightarrow ^{244}_{94}	ext{Pu} + ^{4}_{2}	ext{He}$$

The mass defect is the difference between the mass of the parent nucleus and the sum of the masses of the daughter products:

$$\Delta m = m_{	ext{Cm}} - m_{	ext{Pu}} - m_{	ext{He}}$$

$$\Delta m = 248.072220 - 244.064100 - 4.002603 = 0.005517 	ext{ u}$$

Using Einstein's mass-energy equivalence, we convert this mass defect into energy by multiplying by $931	ext{ MeV/u}$:

$$E_\alpha = 0.005517 	imes 931 = 5.136 	ext{ MeV}$$

### Synthesizing the Total Power Output

Now we have all the pieces of the puzzle. We know the total number of decays per second, the probability of each decay mode, and the energy released by each mode. The total power ($P$) is the sum of the power from fission and the power from $\alpha$-decay:

$$P = (A 	imes 0.08 	imes E_{	ext{fission}}) + (A 	imes 0.92 	imes E_\alpha)$$

Let's substitute our values:

$$P = (10^7 	imes 0.08 	imes 200) + (10^7 	imes 0.92 	imes 5.136) 	ext{ MeV/s}$$

$$P = (16 	imes 10^6) + (4.725 	imes 10^7) 	ext{ MeV/s}$$

$$P = 2.0725 	imes 10^8 	ext{ MeV/s}$$

Finally, we need to convert this power from $	ext{MeV/s}$ to Watts (Joules/s). Since $1	ext{ MeV} = 1.6 	imes 10^{-13}	ext{ J}$, we multiply our result by this conversion factor:

$$P = 2.0725 	imes 10^8 	imes 1.6 	imes 10^{-13} 	ext{ J/s}$$

$$P = 3.316 	imes 10^{-5} 	ext{ W} \approx 3.32 	imes 10^{-5} 	ext{ W}$$

**The final power output of the sample is $3.32 	imes 10^{-5}	ext{ W}$.** Notice how, despite fission only occurring 8% of the time, its massive energy release makes it the dominant contributor to the total power output!

Visualized Solution

$Decay Modes of Curium$

$Rate of Decay (Activity)$

$Calculating Total Activity$

$Energy of α -decay$

$Calculating E_{α}$

$Total Power Output Formula$

$Substituting Values for Power$

$Calculating Power in MeV/s$

$Converting Power to Watts$

$Conclusion$

The Sigma Insight: Radioactivity

The Dual Nature of Curium's Decay

Calculating the Activity

The Energy of Alpha Decay

Synthesizing the Total Power Output

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Decay Modes of Curium

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Rate of Decay (Activity)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating Total Activity

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Energy of α-decay

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating Eα​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Total Power Output Formula

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substituting Values for Power

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating Power in MeV/s

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Converting Power to Watts

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Conclusion