Question #253925

Compute the total binding energy and the. binding energy per nucleon for

(a) 7Li (b) 20Ne (c) 56Fe (d) 235U.


1
Expert's answer
2021-10-20T15:58:11-0400

(a)

Mass Li(7) = 7.016003 amu

Mass of 3 protons =3×1.007825=3.0218294  amu= 3 \times 1.007825 = 3.0218294 \;amu

Mass of 4 neutrons = 4.0346597 amu

Sum of masses of protons and neutrons = 7.056489 amu

Δ=7.056489 -7.016003 = 0.0405 amu

BE=0.0405×931.494  MeV=37.7  MeVBE = 0.0405 \times 931.494 \;MeV \\ = 37.7 \;MeV

Binding energy per nucleon =BEMass  of  atom= \frac{BE}{Mass \;of \; atom}

=37.77=5.38  MeV= \frac{37.7}{7} = 5.38 \;MeV

(b)

Mass Ne(20) = 19.992439

BE=(10×1.007825+10×1.008665)19.992439=0.172461  amu=0.172491×931.494=160.6  MeVBE = (10 \times 1.007825 + 10 \times 1.008665) -19.992439 = 0.172461 \;amu \\ = 0.172491 \times 931.494 = 160.6 \;MeV

Binding energy per nucleon =160.620=8.03  MeV= \frac{160.6}{20} = 8.03 \;MeV

(c)

Mass of Fe(56) = 55.934935 amu

BE=(26×1.007825+30×1.008665)55.934935=0.51344  amu=0.51344×931.494=497.8  MeVBE = (26 \times 1.007825 + 30 \times 1.008665) -55.934935 = 0.51344 \;amu \\ = 0.51344 \times 931.494 = 497.8 \;MeV

Binding energy per nucleon =497.856=8.88  MeV= \frac{497.8}{56} = 8.88 \;MeV

(d)

Mass of U(235) = 235.043943 amu

BE=(93×1.007825+143×1.008665)235.043943=1.915052  amu=1.915052×931.494=1782.9  MeVBE =(93 \times 1.007825 + 143 \times 1.008665) -235.043943 = 1.915052 \;amu \\ = 1.915052 \times 931.494 = 1782.9 \;MeV

Binding energy per nucleon =1792.9235=7.63  MeV= \frac{1792.9}{235} = 7.63 \;MeV


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