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What is the total energy released in the decay \(n \rightarrow p+e^{-}+\bar{\nu}_{e} ?\)

Short Answer

Expert verified
Based on the given step-by-step solution, the short answer is: The total energy released in the decay of a neutron into a proton, an electron, and an electron anti-neutrino is approximately 2.211 × 10⁻¹³ Joules.

Step by step solution

01

Masses of the particles involved

Firstly, we need to know the masses of the particles involved in the decay: neutron, proton, and electron. The electron anti-neutrino's mass can be considered negligible. The masses (in atomic mass units) are as follows: - Mass of neutron (m_n): 1.008665 u - Mass of proton (m_p): 1.007276 u - Mass of electron (m_e): 0.000549 u
02

Conversion of atomic mass units to kilograms

To use the mass-energy equivalency formula, we need to convert the masses of the particles from atomic mass units (u) to kilograms (kg). We can do that using the following conversion factor: 1 u = 1.66053906660 × 10⁻²⁷ kg. Thus, - m_n = 1.008665 u × 1.66053906660 × 10⁻²⁷ kg/u = 1.6749 × 10⁻²⁷ kg - m_p = 1.007276 u × 1.66053906660 × 10⁻²⁷ kg/u = 1.6726 × 10⁻²⁷ kg - m_e = 0.000549 u × 1.66053906660 × 10⁻²⁷ kg/u = 9.1094 × 10⁻³¹ kg
03

Calculate the energy before and after decay

Now, we can calculate the energy before and after the decay using the mass-energy equivalency formula (E = mc²). The energy before the decay is the energy of the neutron, and the energy after the decay is the sum of the energies for the proton, electron, and electron anti-neutrino (which has negligible energy). - Energy before decay (E_n): E_n = m_n × c² = (1.6749 × 10⁻²⁷ kg) × (3 × 10⁸ m/s)² = 1.5054 × 10⁻¹⁰ J - Energy after decay (E_p + E_e): E_p = m_p × c² = (1.6726 × 10⁻²⁷ kg) × (3 × 10⁸ m/s)² = 1.5032 × 10⁻¹⁰ J E_e = m_e × c² = (9.1094 × 10⁻³¹ kg) × (3 × 10⁸ m/s)² = 8.1871 × 10⁻¹⁴ J
04

Calculate the total energy released

Now that we have the energy before and after the decay, we can find the total energy released in the process by subtracting the final energy from the initial energy. Total energy released (ΔE) = E_n - (E_p + E_e) = 1.5054 × 10⁻¹⁰ J - (1.5032 × 10⁻¹⁰ J + 8.1871 × 10⁻¹⁴ J) = 2.211 × 10⁻¹³ J Thus, the total energy released in the decay of a neutron into a proton, electron, and an electron anti-neutrino is approximately 2.211 × 10⁻¹³ Joules.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Mass-Energy Equivalence
One of the most iconic concepts in physics is mass-energy equivalence, encapsulated in Albert Einstein's famous equation, \( E = mc^2 \). This formula shows the relationship between mass \( (m) \) and energy \( (E) \), indicating that mass can be converted into energy and vice versa.

According to this principle, a small amount of mass holds a vast amount of energy. This reveals why nuclear reactions can release extensive amounts of energy, as they involve changes in mass that convert into energy.

In our neutron decay exercise, we use this equation to determine how much energy is released during the decay process. By calculating the difference in mass before and after decay, we can find the energy released process.
Atomic Mass Unit Conversion
Understanding masses in terms like atomic mass units (u) is essential in nuclear physics. It simplifies expressing small particle masses compared to using kilograms.

An atomic mass unit is defined based on the carbon-12 isotope. However, to calculate energy within calculations like mass-energy equivalence, it's crucial to convert these masses into kilograms.

The conversion factor is: \( 1 \, \text{u} = 1.66053906660 \times 10^{-27} \, \text{kg} \). Using this factor helps us transition from atomic mass units to the SI (International System of Units), enabling accurate energy calculations.
Neutron Decay Process
Neutron decay, specifically beta decay, is a fascinating nuclear process where a neutron \((n)\) transforms into a proton \((p)\), an electron \((e^{-})\), and an electron anti-neutrino \((\bar{u}_{e})\).

In this decay, the neutron loses its neutrality and becomes a positively charged proton while emitting an electron and an anti-neutrino. These particles help redistribute energy and momentum to preserve the laws of physics.

The entire process is extremely quick and often occurs in unstable isotopes, leading to more stable configurations. The exercise involves calculating the energy released from such a transformation.
Energy Calculation in Physics
Energy calculations are a cornerstone of physics problems like the neutron decay scenario. They involve determining initial and final energies, helping to understand energy conservation in processes.

Using mass-energy equivalence, we first calculate the neutron's initial energy. After decay, we sum up the energies of the resulting proton and electron. The anti-neutrino's energy in most calculations is negligible.

The final step involves subtracting the total energy after decay from before the decay. This difference represents the energy released during the process, which gives insight into such a nuclear transformation's efficiency and scope.

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Most popular questions from this chapter

What is the average kinetic energy of protons at the center of a star where the temperature is \(1.00 \cdot 10^{7} \mathrm{~K} ?\) What is the average velocity of those protons?

A nuclear reaction of the kind \({ }_{2}^{3} \mathrm{He}+{ }_{6}^{12} \mathrm{C} \rightarrow \mathrm{X}+\alpha\) is called a pick-up nuclear reaction. a) Why is it called a pick-up reaction, that is, what is picked up, what picked it up, and where did it come from? b) What is the resulting nucleus X? c) What is the \(\mathrm{Q}\) -value of this reaction? d) Is this reaction endothermic or exothermic?

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\(^{214} \mathrm{Pb}\) has a half-life of \(26.8 \mathrm{~min}\). How many minutes must elapse for \(90.0 \%\) of a given sample of \({ }^{214} \mathrm{~Pb}\) atoms to decay?

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