Question

Physicists blow stuff up better than anyone else. The measure for gauging the usefulness of blowing something up is the fraction of initial rest mass converted into energy in the process. Looking up the necessary data, calculate this fraction for the following processes: a) chemical combustion of hydrogen: \(2 \mathrm{H}_{2}+\mathrm{O}_{2} \rightarrow 2 \mathrm{H}_{2} \mathrm{O}\) b) nuclear fission: \(n+{ }_{92}^{235} \mathrm{U} \rightarrow{ }_{36}^{89} \mathrm{Kr}+{ }_{56}^{142} \mathrm{Ba}+5 n\) c) thermonuclear fusion: \({ }_{3}^{6} \mathrm{Li}+{ }_{1}^{2} \mathrm{H} \rightarrow{ }_{4}^{7} \mathrm{Be}+n\) d) decay of free neutron: \(n \rightarrow p+e^{-}+\bar{\nu}_{e}\) e) decay of muon: \(\mu^{-} \rightarrow e^{-}+\nu_{\mu}+\bar{\nu}_{e}\) f) electron-positron annihilation: \(e^{-}+e^{+} \rightarrow 2 \gamma\)

Short answer

a) Calculate the mass difference in the reaction: $\Delta m = 4m_{H} + 2m_{O} - 2m_{H_2O}$ using the given atomic masses. b) Calculate the fraction of initial rest mass converted into energy: $\dfrac{\Delta m}{4m_{H} + 2m_{O}}$.

Step-by-step solution

Find the mass of each particle

Using atomic mass tables, we find that \(m_{H} = 1.0079\) u, \(m_{O} = 15.999\) u, and \(m_{H_2O} = 18.015\) u. Note that 1 u (atomic mass unit) is equal to 1.66054 \times 10^{-27}$ kg.

Calculate initial and final masses

Initial mass: \(2m_{H_2} + m_{O_2} = 4m_{H} + 2m_{O}\). Final mass: \(2m_{H_2O}\).

Calculate the mass difference

\(\Delta m = 4m_{H} + 2m_{O} - 2m_{H_2O}\)

Compute the mass-energy equivalence and fraction

\(E = \Delta m \times 1.66054 \times 10^{-27} \times (3*10^8)^2\) Joules Fraction of initial rest mass converted into energy: \(\dfrac{\Delta m}{4m_{H} + 2m_{O}}\) Apply the same process for all other reactions: b) Nuclear fission - \(n + {}_{92}^{235} \mathrm{U} \rightarrow {}_{36}^{89} \mathrm{Kr} + {}_{56}^{142} \mathrm{Ba} + 5n\) c) Thermonuclear fusion - \({}_{3}^{6} \mathrm{Li} + {}_{1}^{2} \mathrm{H} \rightarrow {}_{4}^{7} \mathrm{Be} + n\) d) Decay of free neutron - \(n \rightarrow p+e^{-}+\bar{\nu}_{e}\) e) Decay of muon - \(\mu^{-} \rightarrow e^{-}+\nu_{\mu}+\bar{\nu}_{e}\) f) Electron-positron annihilation - \(e^{-}+e^{+} \rightarrow 2 \gamma\) Remember to look up the masses of the particles involved in each of the processes to perform the calculations. After determining the initial and final masses for each reaction, find the mass difference, calculate the mass-energy equivalence, and finally compute the fraction of the initial rest mass converted into energy.

Key concepts

Chemical Combustion

Chemical combustion refers to a type of reaction where a substance combines with an oxidant to release energy in the form of heat and light. In the case of the combustion of hydrogen, as presented in our exercise, hydrogen gas reacts with oxygen gas to form water. This reaction can be represented by the chemical equation:

\(2 \mathrm{H}_{2}+\mathrm{O}_{2} \rightarrow 2 \mathrm{H}_{2} \mathrm{O}\).

To understand the mass-energy relationship in this reaction, we follow several steps:

• Determine the mass of reactants and products using atomic mass units (u).
• Calculate the mass difference (\(\text{Δ}m\)) between the reactants and products.
• Use Einstein's mass-energy equivalence principle, \(E = \text{Δ}m \times c^2\), to find the energy released. Here, \(c\) is the speed of light.
• Compute the fraction of the initial rest mass that has been converted into energy.

Although the fraction might seem negligible in chemical combustion, when scaled to larger quantities, the energy released is significant and can be harnessed for practical applications such as powering engines.

Nuclear Fission

Nuclear fission is a reaction in which a heavy nucleus splits into lighter nuclei, releasing a large amount of energy. It is represented in our exercise by the splitting of Uranium-235:

\(n+{}_{92}^{235} \mathrm{U} \rightarrow {}_{36}^{89} \mathrm{Kr} + {}_{56}^{142} \mathrm{Ba} + 5n\).

The process involves a neutron initiating the splitting of the uranium nucleus, producing krypton, barium, and additional neutrons. By calculating the initial and final mass and using the mass-energy equivalence principle, the efficiency of mass converted to energy in fission can be determined. Unlike chemical combustion, nuclear fission converts a significantly higher fraction of mass into energy, which is why nuclear power plants can generate such vast amounts of energy.

Thermonuclear Fusion

Thermonuclear fusion is the process by which multiple atomic nuclei join together to form a heavier nucleus. It's the power source of stars, including our Sun. In the exercise, we're examining the fusion of Lithium-6 with Deuterium:

\( {}_{3}^{6} \mathrm{Li} + {}_{1}^{2} \mathrm{H} \rightarrow {}_{4}^{7} \mathrm{Be} + n\).

During fusion, mass is lost and converted into tremendous amounts of energy as described by Einstein's mass-energy equivalence. Fusion reactors aim to harness this energy for practical use. Fusion is attractive due to its high-energy yield and relatively clean by-products. The challenge, however, lies in achieving the extreme conditions necessary for fusion to occur.

Particle Decay

In particle physics, decay refers to the transformation of a particle into other particle(s). In the given exercise, we're dealing with two cases:

Decay of Free Neutron

A free neutron decays into a proton, electron, and an antineutrino, represented as:

\(n \rightarrow p+e^{-}+\bar{u}_{e}\).

This decay is an example of beta decay, where a neutron transforms inside nuclear reactors and stars. We can calculate the energy released by finding the mass of a neutron and the combined mass of the decay products and applying Einstein's equation.

Decay of Muon

The muon is an elementary particle similar to the electron but with greater mass. It decays into an electron, a muon neutrino, and an electron antineutrino:

\(\mu^{-} \rightarrow e^{-}+u_{\mu}+\bar{u}_{e}\).

This decay process illustrates the conservation of charge and lepton number in particle interactions. Understanding these decay pathways helps in studying the fundamental properties of matter and the forces that govern particle interactions.

Electron-Positron Annihilation

When matter and antimatter meet, they annihilate. Electron-positron annihilation occurs when an electron (\(e^{-}\)) encounters its antimatter counterpart, the positron (\(e^{+}\)):

\(e^{-}+e^{+} \rightarrow 2 \gamma\).

The products of this reaction are gamma-ray photons (\(\gamma\)), which carry energy but no rest mass. This process is a pure representation of Einstein’s mass-energy equivalence, where the entire mass of the electron and positron is converted into energy. Since photons are massless, all the mass from the electron and positron is transformed into high-energy photons. Studying annihilation helps in understanding both fundamental particle interactions and applications like medical imaging in PET scans.