A star initially has 10^40 deuterons. It produces energy via the processes 1^2H + 1^2H - 1^3H + p a

Q: Why do we add the two reactions?

Adding the reactions gives us the net overall process. The tritium produced in the first step is immediately consumed in the second step, so it acts as an intermediate. The net reaction shows exactly what goes in and what comes out.

Q: Why do we divide by 3 when calculating the total energy?

The net reaction shows that $3$ deuterons are consumed to release the energy $Q$. Therefore, the total number of reactions that can occur is the total number of deuterons divided by $3$.

Q: How do we convert MeV to Joules?

The conversion factor is $1 \text{ eV} = 1.6 \times 10^{-19} \text{ J}$. Since $1 \text{ MeV} = 10^6 \text{ eV}$, the conversion factor becomes $1 \text{ MeV} = 1.6 \times 10^{-13} \text{ J}$.

Q: What exactly is mass defect?

Mass defect is the difference between the mass of the reactants and the mass of the products in a nuclear reaction. This 'missing' mass is converted into energy according to Einstein's equation $E = mc^2$.

Q: Why is the final answer approximate?

The exact calculated time is $1.29 \times 10^{12} \text{ s}$. The question asks for the 'order of' the time, which means we are looking for the closest power of 10 among the given options, which is $10^{12} \text{ s}$.

Question

A star initially has $10^{40}$ deuterons. It produces energy via the processes $_1^2	ext{H} + _1^2	ext{H} ightarrow _1^3	ext{H} + p$ and $_1^2	ext{H} + _1^3	ext{H} ightarrow _2^4	ext{He} + n$.
If the average power radiated by the star is $10^{16}$ W, the deuteron supply of the star is exhausted in a time of the order of

The mass of the nuclei are as follows:
$	ext{M}(	ext{H}^2) = 2.014 	ext{ amu}$; $	ext{M}(n) = 1.008 	ext{ amu}$;
$	ext{M}(p) = 1.007 	ext{ amu}$; $	ext{M}(	ext{He}^4) = 4.001 	ext{ amu}$.

Options:

(A) $10^6$ s

(B) $10^8$ s

(C) $10^{12}$ s

(D) $10^{16}$ s

Sigma Percentile · Accepted Answer

### The Power of Stars: Nuclear Fusion

Have you ever wondered how stars, like our Sun, keep shining for billions of years? The secret lies in their core, where immense pressure and temperature force light nuclei to fuse together, releasing a staggering amount of energy. In this problem, we are going to calculate the lifespan of a hypothetical star powered by the fusion of deuterons.

### Finding the Net Reaction

The star produces energy through a two-step process:
First, two deuterons fuse to form tritium and a proton:
$$ _1^2	ext{H} + _1^2	ext{H} ightarrow _1^3	ext{H} + p $$

Then, another deuteron fuses with the newly formed tritium to produce helium and a neutron:
$$ _1^2	ext{H} + _1^3	ext{H} ightarrow _2^4	ext{He} + n $$

To find the overall effect, we can simply add these two reactions together. The tritium ($ _1^3	ext{H} $) produced in the first step is consumed in the second step, so it cancels out. We are left with the net reaction:
$$ 3_1^2	ext{H} ightarrow _2^4	ext{He} + n + p $$

This tells us that **three deuterons** are consumed to produce one helium nucleus, one neutron, and one proton.

### The Missing Mass: Calculating Mass Defect

Where does the energy come from? It comes from a tiny amount of mass that goes "missing" during the reaction, converted into energy according to Einstein's famous equation, $E = mc^2$. This missing mass is called the **mass defect** ($\Delta m$).

Let's calculate the mass of the reactants (3 deuterons):
$$ 	ext{Mass of Reactants} = 3 	imes 2.014 	ext{ amu} = 6.042 	ext{ amu} $$

Now, let's calculate the mass of the products (He, n, p):
$$ 	ext{Mass of Products} = 4.001 + 1.008 + 1.007 = 6.016 	ext{ amu} $$

The mass defect is the difference:
$$ \Delta m = 6.042 - 6.016 = 0.026 	ext{ amu} $$

### Converting Mass to Energy

Now, we convert this mass defect into energy. We know that $1 	ext{ amu}$ is equivalent to $931 	ext{ MeV}$ of energy.
$$ Q = 0.026 	imes 931 	ext{ MeV} = 24.206 	ext{ MeV} $$

Since the star's power is given in Watts (Joules per second), we need to convert this energy into Joules. The conversion factor is $1 	ext{ MeV} = 1.6 	imes 10^{-13} 	ext{ J}$.
$$ Q = 24.206 	imes 1.6 	imes 10^{-13} 	ext{ J} \approx 3.87 	imes 10^{-12} 	ext{ J} $$

This is the energy released every time **three deuterons** fuse.

### The Star's Lifespan

The star initially has $10^{40}$ deuterons. Since it takes 3 deuterons to release $Q$ amount of energy, the total energy the star can produce is:
$$ E_{	ext{total}} = \frac{10^{40}}{3} 	imes 3.87 	imes 10^{-12} 	ext{ J} = 1.29 	imes 10^{28} 	ext{ J} $$

We are given that the star radiates power at a rate of $P = 10^{16} 	ext{ W}$ (or $10^{16} 	ext{ J/s}$). Power is the rate of energy consumption ($P = E/t$). Therefore, the time $t$ it takes to exhaust the fuel is:
$$ t = \frac{E_{	ext{total}}}{P} = \frac{1.29 	imes 10^{28}}{10^{16}} 	ext{ s} = 1.29 	imes 10^{12} 	ext{ s} $$

Looking at the options, the time is of the order of **$10^{12} 	ext{ s}$**.

Select Answer:

Visualized Solution

$Net Nuclear Reaction$

$Mass Defect (Δ m)$

$Energy Released (Q)$

$Total Energy for 1 0^{40} Deuterons$

$Time to Exhaust Supply$

The Sigma Insight: Nucleus and Nuclear Reaction

The Power of Stars

Finding the Net Reaction

The Missing Mass

Converting Mass to Energy

The Star's Lifespan

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