Answer to Question #253925 in Atomic and Nuclear Physics for ran

Question

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

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

Answer 1

(a)

Mass Li(7) = 7.016003 amu

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

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

$BE = 0.0405 \times 931.494 \;MeV \\ = 37.7 \;MeV$

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

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

(b)

Mass Ne(20) = 19.992439

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

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

(c)

Mass of Fe(56) = 55.934935 amu

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

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

(d)

Mass of U(235) = 235.043943 amu

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

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