consider the nuclear reaction 42he+73li→105b+10n.what is the energy absorbed or liberated?

the correct equation should look like this

⁴ ₂He + ⁷ ₃Li → ¹⁰ ₅B + ¹n

 to calculate the energy change in this reaction , we need to find the mass difference between the reactant and the product.

 therefore mass on the reactant side is

4.002602+ 6.941=10.943602 amu

and the mass on the product side is

10.811 + 1.008664 =11.819664 amu

 the difference in mass is 10.943602-11.819664 =-0.876amu.

e=mc² , -0.876amu * 3*10⁸= 

converting amu to kg

 we have

e = 1.454632*10^-27 * 3*10⁸ = -4.36* 10^-19j

since the energy is negative , it is obsorbbed, remember the above equation is a fusion, therefore 4.36* 10^-19j  energy is needed to cause the fusion process.