Question 19N.3.hl.TZ0.18

${}_2^4\text{He + }{}_4^8\text{Be}\to {}_6^{12}\text{C}$✔

NOTE: Do not penalize missing atomic numbers.

ALTERNATIVE 1
binding energy per nucleon is larger in carbon-12/product «than beryllium-8 and helium-4/reactants» ✔

difference in «total» binding energy is released «during fusion» ✔

ALTERNATIVE 2
mass of carbon-12/product «nucleus» is less than «the sum of» the masses of helium-4 and beryllium-8 «nuclei»/reactants
OR
two smaller nuclei form a lager nucleus ✔

mass lost/difference is converted to energy «and released»
OR
E = mc² ✔

Δm = «12.000000 amu − (4.002602 amu + 8.005305 amu) =» −0.007907 «amu» ✔
«0.007907 amu × 1.66 × 10⁻²⁷ kg amu⁻¹ =» 1.31 × 10⁻²⁹ «kg» ✔
«E = mc² = 1.31 × 10⁻²⁹ kg × (3.00 × 108 m s⁻¹)² =» 1.18 × 10⁻¹² «J» ✔

NOTE: Accept “0.007907 «amu»”.
Award [2 max] for “7.12 x 10¹⁴ «J»”.
Award [3] for correct final answer.

ALTERNATIVE 1
3 half-lives ✔
0.500 g «of beryllium-8 remain» ✔

ALTERNATIVE 2
$m=4.00{\left(\frac{1}{2}\right)}^{\frac{2.01\times {10}^{-16}}{6.70\times {10}^{-17}}}$ ✔
0.500 g «of beryllium-8 remain» ✔

ALTERNATIVE 3
λ = « $\frac{\ln 2}{6.70\times {10}^{-17}}$ »= 1.03 × 10¹⁶ «s⁻¹» ✔
m = « $4.00{\text{e}}^{-1.03\times {10}^{16}\times 2.01\times {10}^{-16}}\text{ = }$ » 0.500 «g» ✔

NOTE: Award [2] for correct final answer.