Physics question 4 hw 22

Physics
Tutor: None Selected Time limit: 1 Day

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.

Jul 17th, 2015

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1. 

Correct answer: 160.651 MeV. 

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 .

Please let me know if you need any clarification. I'm always happy to answer your questions.
Jul 17th, 2015

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