When 7Li3 nuclei are bombarded by protons, and the resultant nuclei are 8Be4, the emitted particles

Q: Why is a proton written as ${}_1^1\text{H}$?

A standard hydrogen atom consists of one proton and one electron. If you remove the electron, you are left with just the nucleus, which is a single proton. Hence, its atomic number (number of protons) is 1 and its mass number (protons + neutrons) is 1.

Q: What does it mean if the mass number $A$ is zero?

A mass number of zero means the particle does not contain any nucleons (protons or neutrons). It is not a heavy particle like an alpha or beta particle.

Q: Could the emitted particle be a neutrino?

Neutrinos also have zero charge and effectively zero mass number. However, in typical strong nuclear reactions like this one involving nucleon rearrangement, excess energy is usually carried away by gamma photons. Neutrinos are typically associated with weak interactions like beta decay.

Q: Why isn't it an alpha particle?

An alpha particle is a helium nucleus, denoted as ${}_2^4\text{He}$. It has an atomic number of 2 and a mass number of 4. Our calculations strictly showed the unknown particle has $Z=0$ and $A=0$.

Q: Is energy conserved in this reaction?

Yes, total mass-energy is conserved. The slight difference in rest mass between the reactants and products (mass defect) is converted into the kinetic energy of the products and the energy of the emitted gamma photon, according to Einstein's equation $E=mc^2$.

Question

When ${}_3^7	ext{Li}$ nuclei are bombarded by protons, and the resultant nuclei are ${}_4^8	ext{Be}$, the emitted particles will be

Options:

(A) alpha particles

(B) beta particles

(C) gamma photons

(D) neutrons

Sigma Percentile · Accepted Answer

### The Microscopic Collision
Imagine you are observing the microscopic world. We have a Lithium-seven nucleus resting peacefully. Suddenly, it is bombarded by a high-speed proton.

Now, remember that a proton is simply the nucleus of a Hydrogen-one atom, denoted as ${}_1^1	ext{H}$. When these two particles collide, a nuclear reaction takes place. They fuse together to form a new nucleus, Beryllium-eight (${}_4^8	ext{Be}$), and in the process, they emit a mystery particle. Let's call this unknown particle ${}_Z^A	ext{X}$.

### The Golden Rules of Nuclear Physics
To figure out the identity of this mystery particle, we don't need to guess. We just need to apply the golden rules of nuclear physics. In any nuclear reaction like this, two fundamental quantities are strictly conserved.

First, the total atomic number, which represents the total electrical charge, must remain constant before and after the reaction. Second, the total mass number, which is the total count of protons and neutrons, must also be conserved. These two principles will guide us straight to the answer.

### Solving for the Charge
Let's apply the conservation of atomic number first. We look at the subscripts in our nuclear equation:

$$ {}_3^7	ext{Li} + {}_1^1	ext{H} ightarrow {}_4^8	ext{Be} + {}_Z^A	ext{X} $$

On the reactant side, Lithium has an atomic number of $3$, and the proton has an atomic number of $1$. On the product side, Beryllium has an atomic number of $4$. Let's assume our unknown particle X has an atomic number of $Z$. Setting up the equation, we get $3 + 1 = 4 + Z$.

Solving this simple equation gives us $4 = 4 + Z$, which means **$Z = 0$**. This tells us a crucial piece of information: our mystery particle carries absolutely no electrical charge.

### Solving for the Mass
Next, we apply the conservation of mass number. We look at the superscripts in the equation. For the reactants, Lithium has a mass number of $7$, and the proton has a mass number of $1$.

For the products, Beryllium has a mass number of $8$. Let's assume the unknown particle has a mass number of $A$. Our equation becomes $7 + 1 = 8 + A$.

Adding the numbers on the left gives us $8 = 8 + A$. This clearly shows that **$A = 0$**. Our unknown particle has a mass number of zero, meaning it contains no protons or neutrons.

### Unveiling the Mystery Particle
So, what have we discovered? We are looking for a particle with an atomic number of $Z = 0$ and a mass number of $A = 0$.

In the standard model of particle physics, a particle with zero charge and zero rest mass is a **gamma photon** ($\gamma$). It is a packet of pure electromagnetic energy released during the nuclear rearrangement. Therefore, the emitted particles are gamma photons.

Animated Solution for Physics - Atoms and Nuclei: When $_{3}^{7} Li$ nuclei are bombarded by protons, and the resultant nuclei are $_{4}^{8} Be$ , the emitted particles will be

Select Answer:

Visualized Solution

$Nuclear Reaction Setup$

$Conservation Laws$

$Conserving Charge$

$Solving for Z$

$Conserving Mass$

$Solving for A$

$Identifying the Particle$

The Sigma Insight: Nucleus and Nuclear Reaction

The Microscopic Collision Imagine you are observing the microscopic world

The Golden Rules of Nuclear Physics To figure out the identity of this mystery particle, we don't need to guess

Solving for the Charge Let's apply the conservation of atomic number first

Solving for the Mass Next, we apply the conservation of mass number

Unveiling the Mystery Particle

Similar Questions

When boron nucleus $(_{5}^{10} B)$ is bombarded by neutrons, $α$ -particles are emitted. The resulting nucleus is of the element ........... and has the mass number…… .

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

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

List-I

List-II

A nuclear transformation is denoted by $X (n, α) \to_{3}^{7} Li$ . Which of the following is the nucleus of element $X$ ?

Comprehension Passage

The correct statement is

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List-I

List-II

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; } When 37​Li nuclei are bombarded by protons, and the resultant nuclei are 48​Be, the emitted particles will be

Select Answer:

Visualized Solution

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

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

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

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

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

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

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

The Sigma Insight: Nucleus and Nuclear Reaction

The Microscopic Collision Imagine you are observing the microscopic world

The Golden Rules of Nuclear Physics To figure out the identity of this mystery particle, we don't need to guess

Solving for the Charge Let's apply the conservation of atomic number first

Solving for the Mass Next, we apply the conservation of mass number

Unveiling the Mystery Particle

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } When boron nucleus (510​B) is bombarded by neutrons, α-particles are emitted. The resulting nucleus is of the element ........... and has the mass number…… .

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Comprehension Passage

The correct statement is

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List-I

List-II

Animated Solution for Physics - Atoms and Nuclei: When $_{3}^{7} Li$ nuclei are bombarded by protons, and the resultant nuclei are $_{4}^{8} Be$ , the emitted particles will be

$Nuclear Reaction Setup$

$Conservation Laws$

$Conserving Charge$

$Solving for Z$

$Conserving Mass$

$Solving for A$

$Identifying the Particle$

When boron nucleus $(_{5}^{10} B)$ is bombarded by neutrons, $α$ -particles are emitted. The resulting nucleus is of the element ........... and has the mass number…… .

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