Question

Calculate the energy, in megaelectronvolts, released in the nuclear reaction
10 5B + 4 2He? 13 6C + 1 1H. The nuclidic masses are 10 5B = 10.01294u ; 4 2He = 4.002600u ; 13 6C = 13.00335u ; 1 1H = 1.007830u.

Please Show work.

Express your answer with the appropriate units.

Answer 1

Here is what I solved before, please modify the figures as per your question. Please let me know if you have further questions. Ifthis helps then kindly rate 5-stars.

Calculate the energy, in megaelectronvolts, released in the nuclear reaction 6 3Li + 1 0n --> 4 2He + 3 1H. The nuclidic masses are 6 3Li = 6.015130 u; 1 0n = 1.008665 u; 4 2He = 4.002600 u; 3 1H = 3.016040 u.

Express your answer with appropriate units.

yet

e=mc^2

so total mass difference= 6.015130+1.008665-4.002600-3.016040 =0.0052 amu

i'll find out how to convert amu to kgs but it will take sometime.

In the meanwhile:

Whenever you get such a question, it helps to know that 1amu of mass defect releases 931.5 MeV of energy. (where MeV is mega electron volt, 1 electron volt is 1.6 * 10^-19 joules of energy)

i.e. 1amu = 931.5MeV

so we have 0.0052 amu mass defect so

931.5*0.0052 = 4.8019Mev answer