1. Calculate the total binding energy of 20 10Ne. Answer in units of MeV.

2. Calculate the total binding energy of 40 20Ca. Answer in units of MeV.

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1.

Explanation: Let : ZNe−20 = 10 , ANe−20 = 20 , mNe−20 = 19.992435 u ,

mH = 1.007825 u , mn = 1.008665 u , and c 2 = 931.50 MeV/u .

N = A − Z = 20 − 10 = 10

∆m = Z mH + N mn − mNe−20

= 10 (1.007825 u) + 10 (1.008665 u) − 19.992435 u

= 0.172465 u , so Ebind = (0.172465 u)(931.50 MeV/u) = 160.651 MeV .

2.

Correct answer: 342.055 MeV.

Explanation: Let : ZCa−40 = 20 , ACa−40 = 40 , and mCa−40 = 39.962591 u .

N = 40 − 20 = 20

∆m = Z mH + N mn − mCa−40

= 20 (1.007825 u) + 20 (1.008665 u) − 39.962591 u

= 0.367209 u ,

so Ebind = ∆m c2 = (0.367209 u)(931.50 MeV/u) = 342.055 MeV .

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