Nuclear Energy -- Diagnostic Tests

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Radioactive decay is like a population of unstable atoms playing musical chairs, each has a fixed chance of sitting down at any moment.

Question:

The nucleus of helium-4 ($^{4}_{2}\text{He}$) has the following data:

• Mass of proton $= 1.00728\,\text{u}$
• Mass of neutron $= 1.00867\,\text{u}$
• Mass of $^{4}_{2}\text{He}$ nucleus $= 4.00150\,\text{u}$

(a) Calculate the mass defect of $^{4}_{2}\text{He}$ in unified atomic mass units and in MeV/c$^2$.

(b) Calculate the binding energy per nucleon.

(c) Explain why the binding energy per nucleon of $^{4}_{2}\text{He}$ is relatively high for a light nucleus.

Take $1\,\text{u} = 931.5\,\text{MeV}/c^2$.

Solution:

(a) Mass defect: $\Delta m = Zm_p + Nm_n - m_{\text{nucleus}}$

In MeV/c$^2$: $\Delta m = 0.03040 \times 931.5 = 28.3\,\text{MeV}/c^2$

(b) Binding energy $= \Delta m \times 931.5 = 28.3\,\text{MeV}$

Binding energy per nucleon $= 28.3/4 = 7.08\,\text{MeV}$

(c) The binding energy per nucleon of $^{4}_{2}\text{He}$ ($7.08\,\text{MeV}$) is high for a light nucleus because it has equal numbers of protons and neutrons (2 and 2), giving it enhanced stability due to the pairing effect. Nuclei with even numbers of both protons and neutrons (doubly magic) are particularly stable. The binding energy per nucleon curve shows a peak in the region of iron-56 ($8.8\,\text{MeV/nucleon}$), and $^{4}_{2}\text{He}$ is already approaching this value. The high binding energy per nucleon makes $^{4}_{2}\text{He}$ (alpha particles) very tightly bound, which is why alpha decay is a common mode of radioactive decay.

Question:

A typical fission reaction is:

Masses:

• $^{235}\text{U}$: $235.0439\,\text{u}$
• $^{1}\text{n}$: $1.0087\,\text{u}$
• $^{141}\text{Ba}$: $140.9139\,\text{u}$
• $^{92}\text{Kr}$: $91.8973\,\text{u}$

(a) Calculate the energy released per fission event.

(b) Calculate the total energy released when $1.0\,\text{kg}$ of $^{235}\text{U}$ undergoes fission.

(c) Compare this energy with the energy released by burning $1.0\,\text{kg}$ of coal (approximately $3.0 \times 10^7\,\text{J}$).

Take $1\,\text{u} = 931.5\,\text{MeV}/c^2 = 1.661 \times 10^{-27}\,\text{kg}$, $1\,\text{MeV} = 1.60 \times 10^{-13}\,\text{J}$.

Solution:

(a) Total mass before: $m_{\text{before}} = 235.0439 + 1.0087 = 236.0526\,\text{u}$

Total mass after: $m_{\text{after}} = 140.9139 + 91.8973 + 3(1.0087) = 140.9139 + 91.8973 + 3.0261 = 235.8373\,\text{u}$

Mass defect: $\Delta m = 236.0526 - 235.8373 = 0.2153\,\text{u}$

Energy released: $E = 0.2153 \times 931.5 = 200.6\,\text{MeV}$

(b) Number of $^{235}\text{U}$ atoms in $1.0\,\text{kg}$:

Total energy: $E = 2.561 \times 10^{24} \times 200.6 \times 1.60 \times 10^{-13}$

(c) Ratio: $\frac{8.22 \times 10^{13}}{3.0 \times 10^7} = 2.74 \times 10^6$

Nuclear fission releases about 2.7 million times more energy per kilogram than burning coal. This enormous energy density is why nuclear power is so attractive, despite the challenges of waste management and safety.

Question:

The first step of the proton-proton chain in the Sun is:

The dominant overall reaction in the Sun is:

Masses: m_p = 1.00728\,\text{u}$$m_{^{4}\text{He}} = 4.00150\,\text{u}$$m_{e^+} = 0.00055\,\text{u}.

(a) Verify the Q-value of the overall reaction using mass data.

(b) Calculate the energy released per kilogram of hydrogen consumed.

(c) The Sun’s luminosity is $3.85 \times 10^{26}\,\text{W}$. Estimate the mass of hydrogen consumed per second.

Take 1\,\text{u} = 931.5\,\text{MeV}/c^2 = 1.661 \times 10^{-27}\,\text{kg}$$c = 3.00 \times 10^8\,\text{m}\,\text{s}^{-1}.

Solution:

(a) Initial mass: $4m_p = 4 \times 1.00728 = 4.02912\,\text{u}$

Final mass: $m_{^{4}\text{He}} + 2m_{e^+} = 4.00150 + 2(0.00055) = 4.00260\,\text{u}$

Mass defect: $\Delta m = 4.02912 - 4.00260 = 0.02652\,\text{u}$

$Q = 0.02652 \times 931.5 = 24.70\,\text{MeV}$

The stated value is $26.7\,\text{MeV}$ Which includes the annihilation energy of the two positrons with electrons ($2 \times 0.511 = 1.02\,\text{MeV}$). Since the hydrogen atoms include electrons, the net reaction includes $e^+ + e^- \to 2\gamma$ Adding $1.02\,\text{MeV}$:

$24.70 + 1.02 = 25.72\,\text{MeV}$

The remaining difference from $26.7\,\text{MeV}$ is due to the neutrinos carrying away about $0.5\,\text{MeV}$ of kinetic energy on average (which is lost from the Sun). The useful energy deposited in the Sun is about $26.2\,\text{MeV}$ per reaction.

(b) 4 protons produce $26.7\,\text{MeV}$ of energy.

Mass of 4 protons $= 4 \times 1.00728 \times 1.661 \times 10^{-27} = 6.694 \times 10^{-27}\,\text{kg}$

Energy per kg: $E = \frac{26.7 \times 10^6 \times 1.60 \times 10^{-19}}{6.694 \times 10^{-27}} = \frac{4.272 \times 10^{-12}}{6.694 \times 10^{-27}} = 6.38 \times 10^{14}\,\text{J}\,\text{kg}^{-1}$

(c) Mass consumption rate: $\dot{m} = L/E_{\text{per kg}} = 3.85 \times 10^{26}/(6.38 \times 10^{14}) = 6.03 \times 10^{11}\,\text{kg}\,\text{s}^{-1}$

The Sun converts approximately 600 million tonnes of hydrogen to helium every second. Only about $0.7\%$ of the mass is converted to energy (the mass defect fraction), so the actual mass loss rate is $\dot{m} \times 0.007 = 4.2 \times 10^9\,\text{kg}\,\text{s}^{-1}$.

Question:

A nuclear reactor uses $^{235}\text{U}$ fuel. Each fission releases $200\,\text{MeV}$ of energy and produces on average $2.5$ neutrons. The reactor operates at a thermal power of $3000\,\text{MW}$.

(a) Calculate the fission rate (fissions per second).

(b) If the neutron multiplication factor is $k = 1.003$ Calculate the reactor period (the time for the power to double).

(c) Explain the role of control rods and the moderator in maintaining $k = 1$.

Solution:

(a) Fission rate: $\dot{n} = P/(E_{\text{per fission}}) = 3000 \times 10^6/(200 \times 10^6 \times 1.60 \times 10^{-19})$

(b) The reactor period $T$ is related to $k$ by:

Where $\ell$ is the mean neutron generation time. For thermal neutrons in a moderated reactor, $\ell \approx 10^{-4}\,\text{s}$.

For $k = 1.003$: $(k - 1)/k \approx 0.003$

Doubling time: $T = \ell/(k - 1) = 10^{-4}/0.003 = 0.033\,\text{s}$

The power doubles every $33\,\text{ms}$ Which is extremely fast. This is why prompt supercritical excursions are dangerous and why delayed neutrons are essential for controllable reactor operation.

(c) Control rods (made of neutron-absorbing materials like boron or cadmium) are inserted or withdrawn to absorb neutrons and control the reaction rate. Inserting rods reduces $k$ (absorbing more neutrons reduces the multiplication factor); withdrawing rods increases $k$.

The moderator (water, graphite, or heavy water) slows down fast neutrons from fission to thermal energies, where the fission cross-section of $^{235}\text{U}$ is much higher. Without a moderator, most neutrons would escape or be captured without causing further fission, and $k$ would be below 1.

For stable operation at constant power, $k = 1$ exactly (critical state). This is maintained by a negative temperature coefficient: if the reactor gets too hot, thermal expansion and Doppler broadening reduce $k$ Providing automatic negative feedback.

Question:

The following binding energies per nucleon are given:

(a) Calculate the energy released when a $^{235}\text{U}$ nucleus splits into two $^{56}\text{Fe}$ nuclei (approximately).

(b) Calculate the energy released when four $^{1}\text{H}$ nuclei fuse to form $^{4}\text{He}$.

(c) Explain why fusion of very heavy nuclei and fission of very light nuclei do not release energy.

Solution:

(a) For $^{235}\text{U}$: total binding energy $= 235 \times 7.59 = 1783.7\,\text{MeV}$

The fission of $^{235}\text{U}$ does not directly produce two $^{56}\text{Fe}$ nuclei ($2 \times 56 = 112$ nucleons, not $235$). Fission fragments are in the mass range 90—140. For a more realistic calculation, consider the approximate symmetric fission $^{235}\text{U} \to ^{118}\text{Pd} + ^{117}\text{Pd} + n$:

Assuming the products have BE/A $\approx 8.5\,\text{MeV}$ (average for medium-mass nuclei):

$\text{BE}_{\text{products}} = 118 \times 8.5 + 117 \times 8.5 = 2007.5\,\text{MeV}$

$\text{BE}_{\text{reactant}} = 1783.7\,\text{MeV}$

Energy released $\approx 2007.5 - 1783.7 = 223.8\,\text{MeV}$

This is consistent with the approximately $200\,\text{MeV}$ released per fission event. The energy release comes from the increase in binding energy per nucleon (from $7.59$ to about $8.5\,\text{MeV/nucleon}$).

(b) Four $^{1}\text{H}$ nuclei have essentially zero binding energy (single nucleons are unbound).

For $^{4}\text{He}$: total binding energy $= 4 \times 7.07 = 28.28\,\text{MeV}$

Energy released $= 28.28 - 0 = 28.28\,\text{MeV}$

This is the net energy from the pp chain first step through to $^{4}\text{He}$ formation.

(c) The binding energy per nucleon curve has a maximum around $^{56}\text{Fe}$ ($8.79\,\text{MeV/nucleon}$). Nuclei to the left of the peak (light nuclei) can release energy by fusion (combining to form heavier nuclei with higher BE/A). Nuclei to the right of the peak (heavy nuclei) can release energy by fission (splitting into medium-mass nuclei with higher BE/A).

Fusing very heavy nuclei (e.g., two $^{235}\text{U}$ nuclei) would produce a nucleus with BE/A lower than $^{235}\text{U}$ (beyond the peak), so energy would be absorbed, not released. Similarly, fission of very light nuclei (e.g., $^{4}\text{He}$) would produce nuclei with lower BE/A, also absorbing energy.

This is why energy release only occurs when moving toward the peak of the BE/A curve: fusion for light nuclei, fission for heavy nuclei.

Question:

The deuterium-tritium fusion reaction is:

Masses: ^{2}\text{H} = 2.01410\,\text{u}$$^{3}\text{H} = 3.01605\,\text{u}$$^{4}\text{He} = 4.00260\,\text{u}$$^{1}\text{n} = 1.00867\,\text{u}.

(a) Verify the Q-value using mass data.

(b) Calculate the energy released per kilogram of D-T fuel.

(c) The oceans contain approximately $4.6 \times 10^{13}\,\text{kg}$ of deuterium. If all this deuterium were used in D-T fusion (with sufficient tritium), calculate the total energy available and compare it with the world’s annual energy consumption (approximately $6.0 \times 10^{20}\,\text{J}$).

Take 1\,\text{u} = 931.5\,\text{MeV}/c^2$$1\,\text{MeV} = 1.60 \times 10^{-13}\,\text{J}.

Solution:

(a) Initial mass: $2.01410 + 3.01605 = 5.03015\,\text{u}$

Final mass: $4.00260 + 1.00867 = 5.01127\,\text{u}$

Mass defect: $\Delta m = 5.03015 - 5.01127 = 0.01888\,\text{u}$

$Q = 0.01888 \times 931.5 = 17.59\,\text{MeV}$

Confirmed (consistent with $17.6\,\text{MeV}$ given).

(b) Each reaction uses 1 D and 1 T atom.

Mass of one D + one T: $m = (2.01410 + 3.01605) \times 1.661 \times 10^{-27} = 5.03015 \times 1.661 \times 10^{-27} = 8.355 \times 10^{-27}\,\text{kg}$

Energy per reaction: $E = 17.6 \times 10^6 \times 1.60 \times 10^{-13} = 2.816 \times 10^{-12}\,\text{J}$

Energy per kg: $E_{\text{kg}} = 2.816 \times 10^{-12}/8.355 \times 10^{-27} = 3.37 \times 10^{14}\,\text{J}\,\text{kg}^{-1}$

(c) Total energy from $4.6 \times 10^{13}\,\text{kg}$ of deuterium (assuming matching tritium):

$E_{\text{total}} = 4.6 \times 10^{13} \times 3.37 \times 10^{14} = 1.55 \times 10^{28}\,\text{J}$

Years of world energy: $1.55 \times 10^{28}/(6.0 \times 10^{20}) = 2.58 \times 10^7 = 25.8\,\text{million years}$

The deuterium in the oceans could theoretically supply the world’s energy needs for about 26 million years. This highlights the enormous potential of nuclear fusion as a virtually limitless energy source. The challenge is achieving the conditions (temperature $\approx 10^8\,\text{K}$ Confinement, and density) for sustained fusion reactions.

Note: In practice, tritium must be “bred” from lithium using the fusion neutrons, and the achievable efficiency is much lower than $100\%$ But the resource is still effectively limitless on human timescales.

Confusing current and voltage: Current (A) is the flow of charge; voltage (V) is the energy per unit charge. Resistance opposes current, not voltage. Students often say “the resistor uses up the voltage” when they should say “the voltage drops across the resistor.”

Forgetting Ohm’s law applies only to ohmic conductors: = IR$is valid only when resistance is constant (ohmic conductors at constant temperature). For non-ohmic devices (diodes, thermistors, filament lamps), resistance changes with voltage or temperature, so = IR$ gives the resistance at that point, not a constant.

Confusing the conventions for electron flow and conventional current: Conventional current flows from positive to negative (the direction a positive charge would move). Electron flow is from negative to positive (the actual movement of electrons). Most circuit analysis uses conventional current. Using electron flow when conventional current is expected gives reversed directions.

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