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Nuclear Physics: Question 2

Syllabus 23.1

Structured A2 7 marks

A student studying nucleosynthesis models the oxygen-16 nuclide, 816O^{16}_{8}\text{O}, whose nucleus contains 8 protons and 8 neutrons. The mass of a free proton is mp=1.007276 um_p = 1.007276\text{ u}, the mass of a free neutron is mn=1.008665 um_n = 1.008665\text{ u}, and the mass of the assembled oxygen-16 nucleus is mnuc=15.990526 um_{nuc} = 15.990526\text{ u}.

Take 1 u=1.66×10−27 kg=931.5 MeV1\text{ u} = 1.66\times10^{-27}\text{ kg} = 931.5\text{ MeV} and c=3.00×108 m s−1c = 3.00\times10^8\text{ m s}^{-1}.

(a) Calculate the mass defect Δm\Delta m of the oxygen-16 nucleus, in u. [2]

(b) (i) Convert this mass defect to kilograms, and hence use E=c2ΔmE=c^2\Delta m to calculate the binding energy of the nucleus in joules. [2]

(b) (ii) Calculate the binding energy of the nucleus in MeV, using 1 u=931.5 MeV1\text{ u}=931.5\text{ MeV}. [2]

(c) Calculate the binding energy per nucleon of oxygen-16, in MeV. [1]

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Worked solution

Part (a): Mass defect

The nucleus is made of 8 separate protons and 8 separate neutrons, so their total mass if they were unbound would be: 8mp+8mn=8(1.007276)+8(1.008665)=8.058208+8.069320=16.127528 u8m_p+8m_n=8(1.007276)+8(1.008665)=8.058208+8.069320=16.127528\text{ u}

The mass defect is the difference between this total and the actual mass of the bound nucleus: Δm=(8mp+8mn)−mnuc=16.127528−15.990526=0.137002 u\Delta m=(8m_p+8m_n)-m_{nuc}=16.127528-15.990526=0.137002\text{ u}

Recompute as a check, grouping the terms differently: 8(mp+mn)=8(1.007276+1.008665)=8×2.015941=16.127528 u8(m_p+m_n)=8(1.007276+1.008665)=8\times2.015941=16.127528\text{ u}, and 16.127528−15.990526=0.137002 u16.127528-15.990526=0.137002\text{ u}. Both routes agree.

Δm=0.137 u (3 s.f.)\Delta m=\boxed{0.137}\text{ u (3 s.f.)}

Part (b)(i): Binding energy in joules

Convert the mass defect to kilograms using 1 u=1.66×10−27 kg1\text{ u}=1.66\times10^{-27}\text{ kg}: Δm=0.137002×1.66×10−27=2.274×10−28 kg\Delta m=0.137002\times1.66\times10^{-27}=2.274\times10^{-28}\text{ kg}

Now apply E=c2ΔmE=c^2\Delta m: E=(3.00×108)2×2.274×10−28=9.00×1016×2.274×10−28E=(3.00\times10^8)^2\times2.274\times10^{-28}=9.00\times10^{16}\times2.274\times10^{-28}

E=2.047×10−11 JE=2.047\times10^{-11}\text{ J}

Recompute as a check, keeping the powers of ten separate from the start: 9.00×2.274=20.479.00\times2.274=20.47, and 1016×10−28=10−1210^{16}\times10^{-28}=10^{-12}, so E=20.47×10−12=2.047×10−11 JE=20.47\times10^{-12}=2.047\times10^{-11}\text{ J}, consistent.

E=2.05×10−11 J (3 s.f.)E=\boxed{2.05\times10^{-11}}\text{ J (3 s.f.)}

Part (b)(ii): Binding energy in MeV

Using the given conversion 1 u=931.5 MeV1\text{ u}=931.5\text{ MeV} directly on the mass defect found in (a): E=Δm×931.5=0.137002×931.5E=\Delta m\times931.5=0.137002\times931.5

E=0.137002×900+0.137002×31.5=123.30+4.32=127.6 MeVE=0.137002\times900+0.137002\times31.5=123.30+4.32=127.6\text{ MeV}

Recompute as a check by multiplying directly: 0.137×931.5=127.6 MeV0.137\times931.5=127.6\text{ MeV}, matching the split-sum method above.

E=128 MeV (3 s.f.)E=\boxed{128}\text{ MeV (3 s.f.)}

Part (c): Binding energy per nucleon

Oxygen-16 has 1616 nucleons in total (8 protons ++ 8 neutrons), so dividing the full (unrounded) binding energy from (b)(ii) by 16: BE per nucleon=127.616=7.98 MeV\text{BE per nucleon}=\frac{127.6}{16}=7.98\text{ MeV}

Recompute as a check: 16×7.98=127.7 MeV16\times7.98=127.7\text{ MeV}, consistent with the binding energy found above (small rounding only).

BE per nucleon=7.98 MeV (3 s.f.)\text{BE per nucleon}=\boxed{7.98}\text{ MeV (3 s.f.)}

Final answers

  • (a) Δm=0.137 u\Delta m=\boxed{0.137}\text{ u}
  • (b)(i) E=2.05×10−11 JE=\boxed{2.05\times10^{-11}}\text{ J}
  • (b)(ii) E=128 MeVE=\boxed{128}\text{ MeV}
  • (c) Binding energy per nucleon =7.98 MeV=\boxed{7.98}\text{ MeV}