Nuclear Physics -- Diagnostic Tests

flowchart TD
    A[Diag Nuclear Physics] --> B[Key Concepts]
    A --> C[Core Principles]
    A --> D[Practical Applications]
    B --> E[Fundamental definitions]
    C --> F[Design patterns]
    D --> G[Real-world usage]

Question:

A sample contains $1.0 \times 10^{20}$ atoms of isotope A, which decays to isotope B with a half-life of $5$ hours. Isotope B decays to stable isotope C with a half-life of $3$ hours. Initially, there is no B or C. Find (a) the number of A atoms remaining after $15$ hours, (b) the activity of A after $15$ hours, (c) the maximum number of B atoms present at any time, and (d) the approximate time at which B reaches its maximum.

Solution:

(a) A atoms remaining after 15 hours:

Number of half-lives of A: $n = 15/5 = 3$

$N_A = N_0 \times \left(\frac{1}{2}\right)^3 = 1.0 \times 10^{20} \times \frac{1}{8} = 1.25 \times 10^{19}$

(b) Activity of A after 15 hours:

$\lambda_A = \frac{\ln 2}{t_{1/2}} = \frac{0.693}{5 \times 3600} = 3.85 \times 10^{-5} \text{ s}^{-1}$

$A = \lambda_A N_A = 3.85 \times 10^{-5} \times 1.25 \times 10^{19} = 4.81 \times 10^{14} \text{ Bq}$

(c) Maximum number of B atoms:

For a decay chain A $\to$ B $\to$ C, the number of B atoms is:

$N_B(t) = \frac{\lambda_A N_0}{\lambda_B - \lambda_A}\left(e^{-\lambda_A t} - e^{-\lambda_B t}\right)$

$\lambda_A = \frac{0.693}{18000} = 3.85 \times 10^{-5}$ s$^{-1}$

$\lambda_B = \frac{0.693}{10800} = 6.42 \times 10^{-5}$ s$^{-1}$

The maximum of $N_B$ occurs when $\frac{dN_B}{dt} = 0$:

$\lambda_A e^{-\lambda_A t} = \lambda_B e^{-\lambda_B t}$

$e^{(\lambda_B - \lambda_A)t} = \frac{\lambda_B}{\lambda_A} = \frac{6.42}{3.85} = 1.668$

$t_{\max} = \frac{\ln 1.668}{\lambda_B - \lambda_A} = \frac{0.511}{(6.42 - 3.85) \times 10^{-5}} = \frac{0.511}{2.57 \times 10^{-5}} = 19883 \text{ s} = 5.52 \text{ hours}$

(d) Maximum number of B atoms:

$N_B(t_{\max}) = \frac{\lambda_A N_0}{\lambda_B - \lambda_A}\left(e^{-\lambda_A t_{\max}} - e^{-\lambda_B t_{\max}}\right)$

$= \frac{3.85 \times 10^{-5} \times 10^{20}}{2.57 \times 10^{-5}} \times \left(e^{-3.85 \times 10^{-5} \times 19883} - e^{-6.42 \times 10^{-5} \times 19883}\right)$

$= 1.498 \times 10^{20} \times (e^{-0.7654} - e^{-1.2763})$

$= 1.498 \times 10^{20} \times (0.4652 - 0.2790)$

$= 1.498 \times 10^{20} \times 0.1862 = 2.79 \times 10^{19}$

Key misconception: Activity $= \lambda N$ (current number of atoms), NOT $\lambda N_0$ (original number). Many students use $N_0$ instead of $N$ after time has passed. Also, half-life is a statistical property — it applies to large numbers of atoms, not to individual atoms.

Question:

The following data are given:

$1 \text{ u} = 931.5 \text{ MeV}/c^2$, $1 \text{ eV} = 1.6 \times 10^{-19}$ J.

(a) Calculate the binding energy per nucleon of $^{56}\text{Fe}$. (b) Calculate the binding energy per nucleon of $^{235}\text{U}$. (c) For the fission reaction $^{235}\text{U} + n \to ^{141}\text{Ba} + ^{92}\text{Kr} + 3n$Calculate the energy released.

Solution:

(a) Binding energy of $^{56}\text{Fe}$:

$^{56}\text{Fe}$ has $Z = 26$ protons and $N = 30$ neutrons.

Mass of constituent nucleons:

$m_{\text{nucleons}} = 26 \times 1.00728 + 30 \times 1.00867 = 26.18928 + 30.26010 = 56.44938 \text{ u}$

Mass defect:

$\Delta m = 56.44938 - 55.93494 = 0.51444 \text{ u}$

Binding energy:

$BE = 0.51444 \times 931.5 = 479.2 \text{ MeV}$

Binding energy per nucleon:

$\frac{BE}{A} = \frac{479.2}{56} = 8.557 \text{ MeV/nucleon}$

(b) Binding energy of $^{235}\text{U}$:

$Z = 92$, $N = 143$.

$m_{\text{nucleons}} = 92 \times 1.00728 + 143 \times 1.00867 = 92.66976 + 144.23981 = 236.90957 \text{ u}$

$\Delta m = 236.90957 - 235.04393 = 1.86564 \text{ u}$

$BE = 1.86564 \times 931.5 = 1738.1 \text{ MeV}$

$\frac{BE}{A} = \frac{1738.1}{235} = 7.396 \text{ MeV/nucleon}$

(c) Energy released in fission:

Reactants: $^{235}\text{U} + n = 235.04393 + 1.00867 = 236.05260$ u

Products: $^{141}\text{Ba} + ^{92}\text{Kr} + 3n = 140.91390 + 91.92620 + 3(1.00867) = 140.91390 + 91.92620 + 3.02601 = 235.86611$ u

Mass defect:

$\Delta m = 236.05260 - 235.86611 = 0.18649 \text{ u}$

Energy released:

$E = 0.18649 \times 931.5 = 173.7 \text{ MeV}$

Key insight: Iron-56 has the highest binding energy per nucleon of any nucleus, making it the most stable. Energy is released both by fission of heavy nuclei (like U-235, moving toward Fe-56) and by fusion of light nuclei (moving toward Fe-56). The BE/A curve peaks at Fe-56.

Question:

A piece of ancient wood has a $^{14}\text{C}$ activity of $2.5$ decays per minute per gram. Living wood has a $^{14}\text{C}$ activity of $15.3$ decays per minute per gram. The half-life of $^{14}\text{C}$ is $5730$ years. (a) Calculate the age of the wood. (b) If the measurement uncertainty in activity is $\pm 0.3$ decays/min/g, what is the uncertainty in the age? (c) What percentage of the original $^{14}\text{C}$ remains?

Solution:

(a) Age of the wood:

$\frac{A}{A_0} = e^{-\lambda t} = \left(\frac{1}{2}\right)^{t/t_{1/2}}$

$\frac{2.5}{15.3} = \left(\frac{1}{2}\right)^{t/5730}$

$0.1634 = \left(\frac{1}{2}\right)^{t/5730}$

$\log(0.1634) = \frac{t}{5730} \log(0.5)$

$\frac{t}{5730} = \frac{\log 0.1634}{\log 0.5} = \frac{-0.7866}{-0.3010} = 2.613$

$t = 2.613 \times 5730 = 14972 \text{ years}$

(b) Uncertainty in age:

Upper bound (activity $= 2.8$):

$\frac{2.8}{15.3} = 0.1830, \quad \frac{t_u}{5730} = \frac{\log 0.1830}{\log 0.5} = \frac{-0.7375}{-0.3010} = 2.450$

$t_u = 2.450 \times 5730 = 14039 \text{ years}$

Lower bound (activity $= 2.2$):

$\frac{2.2}{15.3} = 0.1438, \quad \frac{t_l}{5730} = \frac{\log 0.1438}{\log 0.5} = \frac{-0.8425}{-0.3010} = 2.799$

$t_l = 2.799 \times 5730 = 16038 \text{ years}$

Uncertainty: approximately $\pm 1000$ years (asymmetric: $-933$ to $+1066$).

(c) Percentage remaining:

$\frac{N}{N_0} = \frac{A}{A_0} = 0.1634 = 16.34\%$

Key insight: Carbon dating becomes increasingly inaccurate for older samples because the remaining activity approaches the background radiation level. For samples older than about $50000$ years ($\sim 9$ half-lives), less than $0.2\%$ of the original $^{14}\text{C}$ remains, making reliable dating nearly impossible.

Question:

A nuclear power station uses $^{235}\text{U}$ as fuel. Each fission of $^{235}\text{U}$ releases approximately $200$ MeV of energy. The power station has a thermal efficiency of $33\%$ and produces $500$ MW of electrical power. (a) Calculate the number of $^{235}\text{U}$ fissions per second. (b) The fuel is enriched to $3\%$ $^{235}\text{U}$ (the rest is $^{238}\text{U}$). Calculate the mass of uranium fuel consumed per day. (c) If the waste heat is discharged into a river with a flow rate of $500$ m$^3$ s$^{-1}$Calculate the temperature rise of the river.

Solution:

(a) Fissions per second:

Thermal power output:

$P_{\text{thermal}} = \frac{P_{\text{electrical}}}{\eta} = \frac{500 \times 10^6}{0.33} = 1.515 \times 10^9 \text{ W}$

Energy per fission: $E = 200 \text{ MeV} = 200 \times 10^6 \times 1.6 \times 10^{-19} = 3.2 \times 10^{-11}$ J

$\text{Fissions per second} = \frac{1.515 \times 10^9}{3.2 \times 10^{-11}} = 4.734 \times 10^{19} \text{ s}^{-1}$

(b) Mass of fuel consumed per day:

$^{235}\text{U}$ atoms fissioned per second $= 4.734 \times 10^{19}$

Per day: $N = 4.734 \times 10^{19} \times 86400 = 4.091 \times 10^{24}$ atoms

Mass of $^{235}\text{U}$ fissioned:

$m_{235} = \frac{N \times 235 \times 1.661 \times 10^{-27}}{1} = 4.091 \times 10^{24} \times 3.903 \times 10^{-25} = 1.597 \text{ kg/day}$

Total uranium fuel (enriched to 3%):

$m_{\text{total}} = \frac{1.597}{0.03} = 53.2 \text{ kg/day}$

(c) Temperature rise of the river:

Waste heat: $P_{\text{waste}} = P_{\text{thermal}} - P_{\text{electrical}} = 1515 - 500 = 1015 \text{ MW}$

$P_{\text{waste}} = \rho_{\text{water}} \times \text{flow rate} \times c_{\text{water}} \times \Delta T$

$1015 \times 10^6 = 1000 \times 500 \times 4200 \times \Delta T$

$\Delta T = \frac{1015 \times 10^6}{2.1 \times 10^9} = 0.483°\text{C}$

Key insight: Nuclear power stations have relatively low thermal efficiency ($30$—$35\%$), meaning about two-thirds of the energy is waste heat. This is a significant environmental consideration for power station siting.

Question:

A stationary $^{226}\text{Ra}$ nucleus undergoes alpha decay to $^{222}\text{Rn}$. The kinetic energy of the alpha particle is $4.78$ MeV. (a) Calculate the recoil kinetic energy of the $^{222}\text{Rn}$ nucleus. (b) Calculate the total energy released in the decay. (c) Explain why the alpha particle carries most of the kinetic energy.

Solution:

(a) Recoil energy of $^{222}\text{Rn}$:

By conservation of momentum (initial momentum $= 0$):

$p_\alpha = -p_{\text{Rn}}$

$m_\alpha v_\alpha = m_{\text{Rn}} v_{\text{Rn}}$

Kinetic energies:

$KE_\alpha = \frac{p^2}{2m_\alpha}, \quad KE_{\text{Rn}} = \frac{p^2}{2m_{\text{Rn}}}$

$\frac{KE_{\text{Rn}}}{KE_\alpha} = \frac{m_\alpha}{m_{\text{Rn}}} = \frac{4}{222} = 0.01802$

$KE_{\text{Rn}} = 4.78 \times 0.01802 = 0.0861 \text{ MeV}$

(b) Total energy released:

$Q = KE_\alpha + KE_{\text{Rn}} = 4.78 + 0.0861 = 4.866 \text{ MeV}$

(c) Why the alpha particle carries most KE:

Since $KE \propto 1/m$ for the same momentum, the lighter particle (alpha, mass $4$ u) carries much more kinetic energy than the heavier daughter nucleus (Rn, mass $222$ u). The ratio is inversely proportional to the mass ratio:

$\frac{KE_\alpha}{KE_{\text{Rn}}} = \frac{m_{\text{Rn}}}{m_\alpha} = \frac{222}{4} = 55.5$

The alpha particle carries about $98.2\%$ of the total kinetic energy released.

Key insight: In alpha decay, both energy AND momentum must be conserved. The lighter alpha particle gets most of the kinetic energy. This is analogous to a gun recoiling when fired — the bullet gets much more KE than the gun, even though they have equal and opposite momenta.

Question:

A positron ($e^+$) and an electron ($e^-$) annihilate, producing two gamma-ray photons. (a) Explain why two photons (not one) are produced. (b) Calculate the wavelength of each photon, assuming the particles are essentially at rest. (c) If the positron has a kinetic energy of $0.5$ MeV before annihilation, calculate the new photon wavelength. (d) In a PET scanner, these photons are detected. Explain how the detection of two photons allows localization of the annihilation event.

Solution:

(a) Why two photons:

Conservation of momentum: if the electron and positron are at rest (or nearly so), their total momentum is zero. A single photon always carries momentum ($p = E/c$), so producing one photon would violate momentum conservation. Two photons travelling in opposite directions can have zero total momentum while carrying away all the energy.

(b) Photon wavelength (particles at rest):

Rest mass energy of electron (or positron): $E_0 = m_e c^2 = 0.511$ MeV

Total energy available: $2 \times 0.511 = 1.022$ MeV

Each photon gets half: $E_\gamma = 0.511$ MeV $= 0.511 \times 10^6 \times 1.6 \times 10^{-19} = 8.176 \times 10^{-14}$ J

$\lambda = \frac{hc}{E_\gamma} = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{8.176 \times 10^{-14}} = \frac{1.989 \times 10^{-25}}{8.176 \times 10^{-14}} = 2.432 \times 10^{-12} \text{ m} = 0.00243 \text{ nm}$

(c) With 0.5 MeV kinetic energy:

Total energy $= 2 \times 0.511 + 0.5 = 1.522$ MeV

Each photon: $E_\gamma = 0.761$ MeV $= 0.761 \times 10^6 \times 1.6 \times 10^{-19} = 1.218 \times 10^{-13}$ J

$\lambda = \frac{hc}{E_\gamma} = \frac{1.989 \times 10^{-25}}{1.218 \times 10^{-13}} = 1.633 \times 10^{-12} \text{ m} = 0.00163 \text{ nm}$

The extra kinetic energy produces shorter-wavelength (higher-energy) photons.

(d) PET scanner localization:

The two photons are emitted in exactly opposite directions (back-to-back, $180°$ apart). A ring of detectors surrounds the patient. When two detectors fire simultaneously (within a very short coincidence time window), the annihilation event must have occurred somewhere along the line connecting the two detectors. By recording many such coincidence events from multiple angles, a computer reconstructs the 3D distribution of the positron-emitting tracer using tomographic algorithms.

Key insight: Mass-energy equivalence ($E = mc^2$) is not just theoretical — it is the operating principle of PET scanners, one of the most important medical imaging technologies. The complete conversion of mass to energy is the most efficient energy release possible.

Confusing mass number with atomic mass: Mass number is the total count of protons and neutrons (integer). Atomic mass is the actual mass in unified atomic mass units (not necessarily an integer). Don’t round atomic mass to get mass number.

Assuming half-life means half the sample decays in that time: Half-life is the time for HALF the remaining atoms to decay. After one half-life, 50% remains; after two, 25%; after three, 12.5%. It’s exponential decay, not linear.

Mixing up alpha, beta, and gamma radiation properties: Alpha is largest/heaviest (stopped by paper), beta is medium (stopped by aluminium), gamma is smallest (requires thick lead). Don’t confuse their penetrating power or ionising ability.

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