Calculate the energy released in the following reaction:
3 Li 6 + 0 n 1 changes to 2 He 4 + 1 H 3.
(mass of 3 Li 6 = 6.015120amu; 1 H 3 = 3.016049amu, 2 He 4 = 4.002604amu, 0 n 1 = 1.008665amu)
Physics Q
2007-06-09 11:23 pm
回答 (1)
2007-06-10 6:01 am
✔ 最佳答案
Total mass of 3 Li 6 and 0 n 1 = 6.015120 + 1.008665 = 7.023785 amuTotal mass of 2 He 4 and 1 H 3 = 4.002604 + 3.016049 = 7.018653 amu
So difference in mass is:
7.023785 - 7.018653 = 0.005132 amu
Using the fugure 1 amu = 931 MeV, we have, the energy released in the reaction is:
0.005132 × 931 = 4.778 MeV
參考: My physics knowledge
收錄日期: 2021-04-23 17:01:50
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