The equation; 4 1H1 - 4He2 2+ + 2e- + 26 MeV represents
(a) beta-decay
(b) gamma-decay
(c) fusion
(

Q: Why are there electrons on the product side of the equation?

Four neutral hydrogen atoms have 4 protons and 4 electrons. They fuse to form a Helium nucleus (2 protons, 2 neutrons) and release 2 positrons. The 2 positrons annihilate with 2 of the original electrons, leaving 2 electrons remaining to balance the overall charge.

Q: Where does the 26 MeV of energy come from?

The combined mass of the four hydrogen nuclei is slightly greater than the mass of the resulting helium nucleus. This difference in mass, known as the 'mass defect', is converted into energy according to Einstein's mass-energy equivalence principle, $E = mc^2$.

Q: Is this the exact reaction that happens in the Sun?

Yes! This equation represents the net result of the proton-proton chain reaction, which is the primary source of energy generation in stars similar to the size of our Sun.

Q: Why is this not considered alpha decay?

Alpha decay involves a single, heavy, unstable nucleus spontaneously emitting an alpha particle (a Helium nucleus) to become more stable. In this equation, multiple light nuclei are actively combining, which defines fusion.

Q: What is the main difference between fusion and fission?

Fusion is the combining of light nuclei into a heavier one, while fission is the splitting of a heavy nucleus into lighter ones. Both processes release energy because the resulting nuclei have a higher binding energy per nucleon.

Question

The equation;
$4~_1^1	ext{H} \longrightarrow ~_2^4	ext{He}^{2+} + 2e^- + 26	ext{ MeV}$ represents

Options:

(A) $\beta$-decay

(B) $\gamma$-decay

(C) fusion

(D) fission

Sigma Percentile · Accepted Answer

Have you ever looked up at the Sun and wondered what powers that massive, glowing sphere of plasma? The answer lies in the very equation we are looking at today. This isn't just a random string of symbols; it's the recipe for starlight.

### Analyzing the Reactants

On the left side of our equation, we have $4~_1^1	ext{H}$. These are four hydrogen nuclei, which are essentially just four individual protons. Hydrogen is the lightest and most abundant element in the universe.

In the extreme temperatures and pressures at the core of a star, these protons are stripped of their electrons and zip around at incredible speeds.

### The Magic of Combination

When these protons collide with enough force to overcome their natural electrostatic repulsion, the strong nuclear force takes over. Through a series of steps known as the proton-proton chain, these four light nuclei are squeezed together.

### The Heavyweight Product

Look at the right side of the equation. The result of this intense cosmic dance is $~_2^4	ext{He}^{2+}$, a single helium nucleus (also known as an alpha particle). Notice that we started with four separate entities and ended up with one heavier entity.

This process of lighter nuclei combining to form a heavier nucleus is the textbook definition of **nuclear fusion**.

### The Energy Payoff

But wait, there's more! The mass of the resulting helium nucleus is actually slightly less than the combined mass of the four original protons. Where did that missing mass go?

It was converted entirely into pure energy, exactly $26	ext{ MeV}$ of it, as dictated by Einstein's famous $E=mc^2$. This is the energy that radiates out from the Sun, warming our planet and making life possible.

So, the next time you feel the warmth of the sun on your face, remember this equation. It's a beautiful example of **nuclear fusion** in action!

Animated Solution for Physics - Atoms and Nuclei: The equation; $4_{1}^{1} H ⟶_{2}^{4} He^{2 +} + 2 e^{-} + 26 MeV$ represents

Select Answer:

Visualized Solution

Analyzing the Reactants

Identifying the Products

Defining Nuclear Fusion

Final Conclusion

The Sigma Insight: Nucleus and Nuclear Reaction

Analyzing the Reactants

The Magic of Combination

The Heavyweight Product

The Energy Payoff

Similar Questions

From the following equations pick out the possible nuclear fusion reactions

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.

Match the nuclear processes given in Column I with the appropriate option(s) in Column II.

List-I

List-II

Binding energy per nucleon versus mass number curve for nuclei is shown in figure. $W$ , $X$ , $Y$ and $Z$ are four nuclei indicated on the curve. The process that would release energy is

Some laws/processes are given in Column I. Match these with the physical phenomena given in Column II.

List-I

List-II

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.

If the binding energy per nucleon in $_{3}^{7} Li$ and $_{2}^{4} He$ nuclei are $5.60 MeV$ and $7.06 MeV$ respectively, then in the reaction $p +_{3}^{7} Li \to 2_{2}^{4} He$ energy of proton must be

A nucleus of mass $M$ emits $γ$ -ray photon of frequency $ν$ . The loss of internal energy by the nucleus is [Take, $c$ is the speed of electromagnetic wave.]

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 ......... .

Animated Solution for Physics - Atoms and Nuclei: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The equation; 4 11​H⟶ 24​He2++2e−+26 MeV represents

Select Answer:

Visualized Solution

Analyzing the Reactants

Identifying the Products

Defining Nuclear Fusion

Final Conclusion

The Sigma Insight: Nucleus and Nuclear Reaction

Analyzing the Reactants

The Magic of Combination

The Heavyweight Product

The Energy Payoff

Similar Questions

From the following equations pick out the possible nuclear fusion reactions

Match the nuclear processes given in Column I with the appropriate option(s) in Column II.

List-I

List-II

Some laws/processes are given in Column I. Match these with the physical phenomena given in Column II.

List-I

List-II

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the binding energy per nucleon in 37​Li and 24​He nuclei are 5.60 MeV and 7.06 MeV respectively, then in the reaction p+37​Li→224​He energy of proton must be

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A nucleus of mass M emits γ-ray photon of frequency ν. The loss of internal energy by the nucleus is [Take, c is the speed of electromagnetic wave.]

Animated Solution for Physics - Atoms and Nuclei: The equation; $4_{1}^{1} H ⟶_{2}^{4} He^{2 +} + 2 e^{-} + 26 MeV$ represents

If the binding energy per nucleon in $_{3}^{7} Li$ and $_{2}^{4} He$ nuclei are $5.60 MeV$ and $7.06 MeV$ respectively, then in the reaction $p +_{3}^{7} Li \to 2_{2}^{4} He$ energy of proton must be

A nucleus of mass $M$ emits $γ$ -ray photon of frequency $ν$ . The loss of internal energy by the nucleus is [Take, $c$ is the speed of electromagnetic wave.]