Question

If 235U{ }^{235} \mathrm{U} captures a neutron to form 236U{ }^{236} \mathrm{U} in its ground state, the energy released is B(236U)B(235U)B\left({ }^{236} \mathrm{U}\right)-B\left({ }^{235} \mathrm{U}\right). (a) Prove this statement. (b) Use the binding-energy formula to estimate the energy released, and compare with the observed value of 6.5MeV6.5 \mathrm{MeV}. (Note: We have assumed here that 236U{ }^{236} \mathrm{U} is formed in its ground state, and the 6.5MeV6.5 \mathrm{MeV} is carried away, by a photon, for example. An important alternative is that 236U{ }^{236} \mathrm{U} can be formed in an excited state, 6.5MeV6.5 \mathrm{MeV} above the ground state. This excitation energy can lead to oscillations that cause the nucleus to fission.)

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(a) The binding energy of 236U^{236}\text{U} is the energy needed to separate all of the 9292 protons and 144144 neutrons from its nucleus. It is energy coming from the difference between the total mass of its individual constituents and the mass of the nucleus as a whole. The binding energy can also be written as:

B(236U)=Sn(92,144)+B(235U)B(^{236}\text{U})=S_n(92,144)+B(^{235}\text{U})

That is, the binding energy is equal to the sum of the energy needed to remove a neutron and the binding energy of what remains in the nucleus.

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