It is proposed to use the nuclear fusion reaction; 1^2H + 1^2H - 2^4He in a nuclear reactor of 200

Q: Why do we multiply the number of reactions by 2 to get the number of deuterium nuclei?

Look closely at the balanced fusion equation: $_1^2\text{H} + _1^2\text{H} \rightarrow _2^4\text{He}$. Each single fusion event requires *two* deuterium nuclei to collide and fuse. So, if you need $N$ reactions to produce the required energy, you will consume $2N$ deuterium nuclei.

Q: Why do we use $931.48 \text{ MeV}$ instead of $E=mc^2$ directly with kilograms?

You absolutely could use $E=mc^2$! However, converting atomic mass units (amu) to kilograms and then multiplying by $c^2$ is tedious. The factor $931.48 \text{ MeV/amu}$ is a pre-calculated shortcut derived directly from $E=mc^2$, making nuclear calculations much faster and less prone to arithmetic errors.

Q: Why do we need to convert the energy from MeV to Joules?

The reactor's power is given in Megawatts (MW), which is Megajoules per second ($10^6 \text{ J/s}$). To relate the microscopic energy released per fusion to the macroscopic power output of the reactor, both must be in the same standard SI unit: Joules.

Q: What does the 25% efficiency actually mean in this context?

In any power plant, a lot of energy is lost as waste heat (due to the laws of thermodynamics). A 25% efficiency means that for every 100 Joules of energy released by the fusion reactions, only 25 Joules are successfully converted into usable electrical energy. The rest is lost.

Q: Why is the molar mass of deuterium $2 \text{ g/mol}$?

A deuterium nucleus (a deuteron) consists of one proton and one neutron, giving it an atomic mass number of 2. Therefore, one mole of deuterium atoms has a mass of approximately 2 grams.

Question

It is proposed to use the nuclear fusion reaction;
$$_{1}^{2}	ext{H} + _{1}^{2}	ext{H} ightarrow _{2}^{4}	ext{He}$$
in a nuclear reactor of $200	ext{ MW}$ rating. If the energy from the above reaction is used with a $25$ per cent efficiency in the reactor, how many grams of deuterium fuel will be needed per day? (The masses of $_{1}^{2}	ext{H}$ and $_{2}^{4}	ext{He}$ are $2.0141$ atomic mass units and $4.0026$ atomic mass units respectively.)

Sigma Percentile · Accepted Answer

Imagine a massive nuclear reactor. Inside its core, a fascinating process is happening: nuclear fusion. Two deuterium nuclei are colliding and fusing together to form a single helium nucleus, releasing a tremendous amount of energy in the process. Our goal is to find out exactly how many grams of deuterium this reactor consumes every single day to maintain a $200	ext{ MW}$ output at $25\%$ efficiency.

### The Missing Mass: Calculating Mass Defect

To find the energy released in one single fusion event, we first need to calculate the mass defect. The mass defect is simply the difference between the total mass of our reactants, which are the two deuterons, and the mass of our product, the helium nucleus.

Let's plug in the numbers. We have two deuterium nuclei, each with a mass of $2.0141	ext{ amu}$. From this total, we subtract the mass of the helium nucleus, which is $4.0026	ext{ amu}$.

$$\Delta m = 2(2.0141) - 4.0026$$

When we do the math, we find that the mass defect is $0.0256	ext{ amu}$. This tiny amount of missing mass is what gets converted into pure energy!

### Unleashing the Energy

Now, how do we convert this mass defect into energy? We use the standard conversion factor: $1	ext{ amu}$ is equivalent to $931.48	ext{ MeV}$ of energy.

Multiplying our mass defect by this factor, we get the total energy released per fusion, which is $23.85	ext{ MeV}$. To make this useful for our macroscopic reactor calculations, let's convert this into Joules. Multiplying by $1.6 	imes 10^{-13}$, we get $3.82 	imes 10^{-12}	ext{ J}$.

But wait, there is a catch here. The reactor is not perfectly efficient. It only captures $25\%$ of this released energy to produce useful power. So, we need to find the actual usable energy per fusion.

Taking $25\%$ of our total energy gives us $0.25 	imes 3.82 	imes 10^{-12}$. This results in $9.55 	imes 10^{-13}	ext{ J}$ of usable energy from every single fusion reaction.

### The Reactor's Daily Appetite

Now, let's shift our focus to the reactor's daily output. The reactor operates at a power of $200	ext{ MW}$. To find the total energy required in one full day, we multiply the power by the total time in seconds.

$200	ext{ MW}$ is $200 	imes 10^6	ext{ W}$. A day has $24$ hours, and each hour has $3600$ seconds. Multiplying these together, we find the reactor needs a massive $1.728 	imes 10^{13}	ext{ J}$ of energy every day!

$$E_{	ext{total}} = (200 	imes 10^6) 	imes (24 	imes 3600) = 1.728 	imes 10^{13}	ext{ J}$$

### Counting the Nuclei

So, how many deuterium nuclei do we need to produce this much energy? We simply divide the total daily energy by the usable energy per fusion. But remember, each fusion reaction consumes **two** deuterium nuclei, so we must multiply this ratio by two.

$$n = 2 	imes \frac{1.728 	imes 10^{13}}{9.55 	imes 10^{-13}}$$

Plugging in our values, we get $0.362 	imes 10^{26}$ deuterium nuclei needed per day.

### The Final Weigh-In

We are almost there! We have the number of nuclei, but the question asks for the mass in grams. To convert the number of particles to mass, we divide by Avogadro's number to get the number of moles, and then multiply by the molar mass of deuterium, which is $2	ext{ g/mol}$.

$$	ext{Mass} = \frac{0.362 	imes 10^{26}}{6.02 	imes 10^{23}} 	imes 2$$

Dividing $0.362 	imes 10^{26}$ by $6.02 	imes 10^{23}$, and multiplying by $2$, we get exactly **$120.26	ext{ g}$**. That's our final answer!

Just think about that for a second. A massive $200	ext{ MW}$ power plant running for an entire day on just about $120$ grams of fuel! That is the incredible power of nuclear fusion.

Enter Numerical Value:

Visualized Solution

$Nuclear Fusion Reactor$

$Mass Defect (Δ m)$

$Substituting Masses$

$Calculating Δ m$

$Energy Released (Δ E)$

$Calculating Δ E$

$Usable Energy per Fusion$

$Calculating Usable Energy$

$Total Energy Required per Day$

$Calculating E_{total}$

$Number of Deuterium Nuclei (n)$

$Calculating n$

$Mass of Deuterium Required$

$Final Calculation$

$Conclusion$

The Sigma Insight: Nucleus and Nuclear Reaction

The Missing Mass

Unleashing the Energy

The Reactor's Daily Appetite

Counting the Nuclei

The Final Weigh-In

Similar Questions

In a nuclear reactor $^{235} U$ undergoes fission liberating $200 MeV$ of energy. The reactor has a $10%$ efficiency and produces $1000 MW$ power. If the reactor is to function for $10 yr$ , find the total mass of uranium required.

Consider the reaction : $_{1}^{2} H +_{1}^{2} H =_{2}^{4} He + Q$ . Mass of the deuterium atom = 2.0141 u. Mass of helium atom = 4.0024 u. This is a nuclear ........ reaction in which the energy $Q$ released is ...... MeV.

In a reactor, $2 kg$ of $_{92} U^{235}$ fuel is fully used up in $30 days$ . The energy released per fission is $200 MeV$ . Given that, the Avogadro number, $N = 6.023 \times 1 0^{26} kmol^{- 1}$ and $1 eV = 1.6 \times 1 0^{- 19} J$ . The power output of the reactor is close to

The binding energies per nucleon for deuteron ( $_{1} H^{2}$ ) and helium ( $_{2} He^{4}$ ) are $1.1 MeV$ and $7.0 MeV$ respectively. The energy released when two deuterons fuse to form a helium nucleus ( $_{2} He^{4}$ ) is ......... .

The binding energy per nucleon of deuteron ( $_{1}^{2} H$ ) and helium nucleus ( $_{2}^{4} He$ ) is $1.1 MeV$ and $7 MeV$ respectively. If two deuteron nuclei react to form a single helium nucleus, then the energy released is

Consider the nuclear fission $^{20} Ne ⟶ 2^{4} He +^{12} C$ Given that the binding energy/nucleon of $^{20} Ne$ , $^{4} He$ and $^{12} C$ are respectively, $8.03 MeV$ , $7.07 MeV$ and $7.86 MeV$ , identify the correct statement.

You are given that mass of $_{3}^{7} Li = 7.0160 u$ , mass of $_{2}^{4} He = 4.0026 u$ and mass of $_{1}^{1} H = 1.0079 u$ . When $20 g$ of $_{3}^{7} Li$ is converted into $_{2}^{4} He$ by proton capture, the energy liberated (in $kWh$ ), is [Mass of nucleon $= 1 GeV/c^{2}$ ]

If a star can convert all the He nuclei completely into oxygen nuclei. The energy released per oxygen nuclei is : (Mass of the helium nucleus is $4.0026 amu$ and mass of oxygen nucleus is $15.9994 amu$ )

Enter Numerical Value:

Visualized Solution

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Mass Defect (Δm)

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Energy Released (ΔE)

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

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Total Energy Required per Day

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Number of Deuterium Nuclei (n)

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

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

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

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