Nuclear & Particle Physics - Binding Energy

The ²³⁸U nucleus has a binding energy of about 7.6 MeV per nucleon. If the nucleus were to fission into two equal fragments, each would have a kinetic energy of just over 100 MeV. From this it can be concluded that

A. ²³⁸U cannot fission spontaneously
B. ²³⁸U has a large neutron excess
C. nuclei near A = 120 have masses greater than half that of ²³⁸U
D. nuclei near A = 120 must be bound by about 6.7 MeV/nucleon
E. nuclei near A = 120 must be bound by about 8.5 MeV/nucleon

(GR0177 #67)

Solution:

(A) FALSE.
²³⁸U is a heavy element, it can fission spontaneously.

(B) FALSE.
It is not related to binding energy

(C) FALSE.
Total mass of nucleus is always less than the sum of the masses of its individual nucleons.

(D) FALSE.
Nuclei near A = 120 must be bound by energy greater than A = 238 (must be more than 7.6 MeV per nucleon)

(E) TRUE.
8.5 MeV/nucleon is more than 7.6 MeV per nucleon.

Answer: E

Notes:

To calculate the binding energy per nucleon:

²³⁸U → A = 238 ≈ 240
²³⁸U nucleus were to fission into two equal fragments: A₁ = A₂ = A' = 240/2 = 120

Initial binding energy: E_i = 240 × 7.6 MeV

Final binding energy: E_f = 120 E + 120 E = 240 E 

The kinetic energy of the two fragments is the difference in binding energy between the initial and final state nuclei:



Thus, nuclei near A = 120 must be bound by about 8.5 MeV/nucleon

The graph also shows that for A = 120, binding energy is about 8.5 MeV/nucleon

Source: hyperphysics.phy-astr.gsu.edu