Level 4 — ApplicationNuclear & Radiochemistry

Nuclear & Radiochemistry

60 marksprintable — key stays hidden on paper

Time: 60 minutes Total marks: 60

Use: 1 u = 931.5 MeV, $N_A = 6.022\times10^{23}$ , $c = 3.00\times10^8$ m/s. Show all reasoning.

Q1. Binding energy & stability (12 marks)

A newly reported isotope $^{48}_{20}\text{Ca}$ has an atomic mass of 47.95253 u. Masses: $^1_1\text{H} = 1.00783$ u, neutron $= 1.00867$ u.

(a) Calculate the total binding energy and the binding energy per nucleon (in MeV). (6)

(b) $^{48}\text{Ca}$ is unusually stable for its high N/Z ratio. Referring to both proton and neutron numbers, give the specific structural reason. (3)

(c) Predict the most likely decay mode of $^{45}_{20}\text{Ca}$ (which is unstable) and justify from its N/Z relative to the stable $^{40}\text{Ca}$ . (3)

Q2. Decay kinetics — mixed source (14 marks)

A sealed sample initially contains $2.00\times10^{18}$ atoms of isotope A ( $t_{1/2}=8.0$ h) and $5.00\times10^{17}$ atoms of a different isotope B ( $t_{1/2}=24.0$ h). Both decay independently to stable products.

(a) Compute the decay constant (per hour) and initial activity (in Bq) of each isotope. (6)

(b) Find the total activity of the sample after 24.0 h. (5)

(c) After how many hours does isotope B's activity first exceed isotope A's activity? Give an exact expression then a numerical value. (3)

Q3. Nuclear reaction Q-value & threshold (12 marks)

Consider the reaction used in a neutron generator: $^2_1\text{H} + ^3_1\text{H} \rightarrow ^4_2\text{He} + ^1_0\text{n}$ Masses: $^2\text{H}=2.01410$ u, $^3\text{H}=3.01605$ u, $^4\text{He}=4.00260$ u, $n=1.00867$ u.

(a) Calculate the Q-value in MeV and state whether the reaction is exo- or endoergic. (5)

(b) The emitted neutron carries most of this energy. Given the neutron receives 14.1 MeV, compute its speed (non-relativistic; $m_n=1.675\times10^{-27}$ kg). (4)

(c) Explain physically why the D–T reaction still requires a high ignition temperature despite being strongly exoergic. (3)

Q4. Applications — dating & medicine (12 marks)

(a) A wooden artefact gives a $^{14}\text{C}$ activity of 9.2 disintegrations per minute per gram of carbon; living material gives 15.3 dpm/g. Taking $t_{1/2}(^{14}\text{C})=5730$ yr, calculate the age of the artefact. (5)

(b) $^{99m}\text{Tc}$ ( $t_{1/2}=6.0$ h) is preferred for imaging over $^{131}\text{I}$ ( $t_{1/2}=8.0$ d). Give two nuclear/physical reasons why $^{99m}\text{Tc}$ is the safer imaging choice. (4)

Q5. Radiation safety & shielding (10 marks)

(a) A gamma source's intensity is reduced to 25% after passing through 6.0 cm of lead. Calculate the linear attenuation coefficient $\mu$ (cm⁻¹) and the half-value thickness. (5)

(b) A worker absorbs 0.40 Gy of alpha radiation (radiation weighting factor $w_R=20$ ). Compute the equivalent dose in Sv, and explain why the same 0.40 Gy of gamma ( $w_R=1$ ) is far less biologically damaging per unit absorbed dose. (5)

Answer keyMark scheme & solutions

Q1 (12)

(a) Mass defect using 20 H atoms (accounts for electrons) + 28 neutrons: $\Delta m = 20(1.00783) + 28(1.00867) - 47.95253$ $= 20.15660 + 28.24276 - 47.95253 = 0.44683\text{ u}$ (2 marks setup, 1 mark value) $BE = 0.44683 \times 931.5 = 416.2\text{ MeV}$ (2) $BE/A = 416.2/48 = 8.67\text{ MeV/nucleon}$ (1)

(b) Both Z = 20 and N = 28 are magic numbers → doubly magic nucleus, closed proton and neutron shells give extra stability. (3; 1 for Z=20 magic, 1 for N=28 magic, 1 for doubly-magic)

(c) $^{45}\text{Ca}$ has N = 25, Z = 20 (N/Z = 1.25), higher neutron excess than stable $^{40}\text{Ca}$ (N/Z = 1). To reduce neutrons it undergoes β⁻ decay (n → p + e⁻ + $\bar\nu$ ), moving toward the stability line. (3; 1 mode, 2 justification)

Q2 (14)

(a) $\lambda_A = \ln2/8.0 = 0.08664\ \text{h}^{-1}$ ; $\lambda_B = \ln2/24.0 = 0.02888\ \text{h}^{-1}$ (2)

Convert to per second for Bq: $\lambda/3600$ . $A_{0,A} = \lambda_A N_A = (0.08664/3600)(2.00\times10^{18}) = 4.81\times10^{13}$ Bq (2) $A_{0,B} = (0.02888/3600)(5.00\times10^{17}) = 4.01\times10^{12}$ Bq (2)

(b) After 24 h: A elapsed = 3 half-lives → $A_A = 4.81\times10^{13}/2^3 = 6.01\times10^{12}$ Bq (2) B elapsed = 1 half-life → $A_B = 4.01\times10^{12}/2 = 2.01\times10^{12}$ Bq (2) Total $= 8.02\times10^{12}$ Bq (1)

(c) Set $A_B(t) = A_A(t)$ : $A_{0,B}e^{-\lambda_B t} = A_{0,A}e^{-\lambda_A t}$ $t = \frac{\ln(A_{0,A}/A_{0,B})}{\lambda_A - \lambda_B} = \frac{\ln(4.81\times10^{13}/4.01\times10^{12})}{0.08664 - 0.02888}$ (2) $= \frac{\ln(12.0)}{0.05776} = \frac{2.485}{0.05776} = 43.0\text{ h}$ B first exceeds A just after ~43 h. (1)

Q3 (12)

(a) $\Delta m = (2.01410 + 3.01605) - (4.00260 + 1.00867)$ $= 5.03015 - 5.01127 = 0.01888$ u (3) $Q = 0.01888 \times 931.5 = 17.6\text{ MeV}$ → exoergic (positive Q). (2)

(b) $E = \tfrac12 m v^2$ , $E = 14.1\text{ MeV} = 14.1\times10^6\times1.602\times10^{-19} = 2.259\times10^{-12}$ J (1) $v = \sqrt{\frac{2E}{m}} = \sqrt{\frac{2(2.259\times10^{-12})}{1.675\times10^{-27}}} = \sqrt{2.697\times10^{15}}$ (2) $= 5.19\times10^7\text{ m/s}$ (1)

(c) Both nuclei are positively charged; they must overcome the Coulomb barrier to approach within range of the strong force. This requires high kinetic energy → high temperature (~ $10^8$ K) so nuclei have enough thermal energy to tunnel/fuse. Q-value is released only after fusion occurs. (3)

Q4 (12)

(a) $t = \frac{t_{1/2}}{\ln2}\ln\frac{A_0}{A} = \frac{5730}{0.6931}\ln\frac{15.3}{9.2}$ (3) $= 8266 \times \ln(1.663) = 8266 \times 0.5087 = 4205\text{ yr}$ ≈ 4.2×10³ yr (2)

(b) Any two: (2 each)

Short half-life (6 h) → rapid clearance, low cumulative dose to patient.
Emits pure γ (140 keV), no particulate (α/β) radiation → good imaging penetration but minimal tissue ionisation damage; I-131 emits β⁻ which deposits more damaging dose.
γ energy ideal for detection while minimising absorbed dose.

(c) 18 h = 3 half-lives → $600/2^3 = 75\text{ MBq}$ . (3)

Q5 (10)

(a) $I/I_0 = e^{-\mu x}$ : $0.25 = e^{-\mu(6.0)}$ $\mu = -\ln(0.25)/6.0 = 1.386/6.0 = 0.231\ \text{cm}^{-1}$ (3) Half-value thickness $x_{1/2} = \ln2/\mu = 0.6931/0.231 = 3.0\text{ cm}$ (2)

(b) $H = w_R \times D = 20 \times 0.40 = 8.0\text{ Sv}$ (2) Gamma: $H = 1 \times 0.40 = 0.40$ Sv. Alpha particles are heavy, highly charged → dense ionisation (high LET) along short tracks, causing concentrated, hard-to-repair damage; gamma deposits energy sparsely (low LET), hence $w_R$ 20× lower. (3)

[
{"claim":"Q1a BE/A ≈ 8.67 MeV","code":"dm=20*1.00783+28*1.00867-47.95253; BE=dm*931.5; result=abs(BE/48-8.67)<0.02"},
{"claim":"Q2c crossover time ≈ 43.0 h","code":"import sympy as sp; lA=sp.log(2)/8; lB=sp.log(2)/24; A0A=(lA/3600)*2e18; A0B=(lB/3600)*5e17; t=sp.log(A0A/A0B)/(lA-lB); result=abs(float(t)-43.0)<0.5"},
{"claim":"Q3a Q-value ≈ 17.6 MeV","code":"dm=(2.01410+3.01605)-(4.00260+1.00867); Q=dm*931.5; result=abs(Q-17.6)<0.1"},
{"claim":"Q4a age ≈ 4205 yr","code":"import sympy as sp; t=(5730/sp.log(2))*sp.log(15.3/9.2); result=abs(float(t)-4205)<10"},
{"claim":"Q5a mu ≈ 0.231, HVT ≈ 3.0","code":"import sympy as sp; mu=-sp.log(0.25)/6.0; hvt=sp.log(2)/mu; result=abs(float(mu)-0.231)<0.005 and abs(float(hvt)-3.0)<0.05"}
]