The binding energies per nucleon for deuteron (1H2) and helium (2He4) are 1.1 MeV and 7.0 MeV respec

Q: Why do we subtract the reactant binding energy from the product binding energy to find the energy released?

Think of binding energy as a measure of stability. A higher binding energy means the nucleus is in a lower, more stable energy state. When reactants with lower binding energy form products with higher binding energy, the system drops to a lower energy state, and the difference in energy is released into the surroundings.

Q: What exactly is binding energy per nucleon?

It is the total binding energy of a nucleus divided by its mass number (the total number of protons and neutrons). It represents the average energy required to remove a single nucleon from the nucleus, and it is a great indicator of how stable a nucleus is.

Q: Why does fusion release energy if we are building a heavier nucleus?

Fusion releases energy when light nuclei (lighter than iron) combine. This is because the resulting heavier nucleus has a higher binding energy per nucleon than the lighter reactants. The nucleons become more tightly bound, and the 'lost' mass is converted into energy.

Q: Is this the exact reaction that powers the sun?

The sun primarily uses a process called the proton-proton chain, where hydrogen nuclei (protons) fuse to eventually form helium. However, the fusion of deuterons is a crucial intermediate step in that chain and in many artificial fusion reactor designs.

Q: Why is the binding energy per nucleon of Helium so much higher than that of Deuteron?

Helium-4 is what physicists call a 'magic number' nucleus. It has 2 protons and 2 neutrons, which completely fill its lowest nuclear energy shells. This closed-shell structure makes it exceptionally tightly bound and stable compared to the loosely bound deuteron.

Question

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

Sigma Percentile · Accepted Answer

### The Cosmic Engine
Have you ever looked up at the night sky and wondered what powers those brilliant points of light? The answer lies in the very heart of atoms, in a process called nuclear fusion.

This is the exact same process we are going to explore in this problem. We are looking at the fusion of two deuteron nuclei to form a single helium nucleus.

### Analyzing the Setup
Let us break down the players in our cosmic drama. We start with two deuterons. A deuteron is the nucleus of deuterium, a heavy isotope of hydrogen, consisting of one proton and one neutron.

When these two deuterons collide with enough energy, they overcome their electrostatic repulsion and fuse together. The result is a helium-4 nucleus, also known as an alpha particle, which contains two protons and two neutrons.

But here is the magical part: the mass of the resulting helium nucleus is slightly less than the combined mass of the two original deuterons. Where did this missing mass go?

According to Einstein's famous equation, it is converted into pure energy!

### The Master Equation
To calculate this released energy, we do not need to measure the masses directly. We can use the concept of binding energy.

**Binding energy** is the energy required to completely disassemble a nucleus into its constituent protons and neutrons. Conversely, it is the energy released when those nucleons bind together to form the nucleus.

The problem gives us the **binding energy per nucleon**. This is a measure of how tightly bound the nucleus is. A higher binding energy per nucleon means a more stable nucleus.

The total energy released in a fusion reaction is simply the difference between the total binding energy of the products and the total binding energy of the reactants.

$$ \Delta E = 	ext{Total BE of Products} - 	ext{Total BE of Reactants} $$

### Calculating the Reactant Energy
Let us calculate the total binding energy of our reactants. We have two deuterons.

Each deuteron has $2$ nucleons. The problem states that the binding energy per nucleon for a deuteron is $1.1 	ext{ MeV}$.

So, the binding energy of one deuteron is $2 	imes 1.1 	ext{ MeV} = 2.2 	ext{ MeV}$.

Since we have two deuterons fusing, the total binding energy of the reactants is:

$$ 	ext{BE}_{	ext{reactants}} = 2 	imes 2.2 	ext{ MeV} = 4.4 	ext{ MeV} $$

### Calculating the Product Energy
Now, let us look at our product, the helium nucleus.

A helium-4 nucleus has $4$ nucleons. Its binding energy per nucleon is given as $7.0 	ext{ MeV}$. This high value indicates that helium-4 is an exceptionally stable nucleus.

The total binding energy of the helium nucleus is:

$$ 	ext{BE}_{	ext{product}} = 4 	imes 7.0 	ext{ MeV} = 28.0 	ext{ MeV} $$

### Final Calculation
We are now ready for the final step. We subtract the total binding energy of the reactants from the total binding energy of the product.

$$ \Delta E = 28.0 	ext{ MeV} - 4.4 	ext{ MeV} $$

$$ \Delta E = 23.6 	ext{ MeV} $$

This is the energy released in just one single fusion event!

While **$23.6 	ext{ MeV}$** might seem like a tiny amount of energy on a macroscopic scale, remember that a single gram of deuterium contains roughly $3 	imes 10^{23}$ atoms. If all of them fused, the energy released would be astronomical.

This is the very principle that scientists and engineers are trying to harness in fusion reactors on Earth, hoping to provide a clean, virtually limitless source of energy for the future.

Animated Solution for Physics - Atoms and Nuclei: 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 ......... .

Enter Numerical Value:

Visualized Solution

$Nuclear Fusion Reaction$

$Energy Released Formula$

$Binding Energy of Reactants$

$Binding Energy of Product$

$Final Calculation$

$The Way Forward$

The Sigma Insight: Nucleus and Nuclear Reaction

The Cosmic Engine

Analyzing the Setup Let us break down the players in our cosmic drama

The Master Equation To calculate this released energy, we do not need to measure the masses directly

Calculating the Reactant Energy Let us calculate the total binding energy of our reactants

Calculating the Product Energy

Final Calculation We are now ready for the final step

Similar Questions

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Comprehension Passage

The correct statement is

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