It has been estimated that the total power radiated by the sun is 3.8 xx 10^(26) J per second . The source of energy of stars is a thermonucleur fission reaction . Energy released in the process of fusion is due to mass defect . Q = Delta mc^(2) In a nuclear reaction , ""_(1)^(2) H + ""_(1)^(2) H to ""_(2)^(3) He + ""_(0)^(1)n . if the masses of ""_(1)^(2) H & ""_(2)^(3) He are 2.014741 amu and 3.016977 amu respectively . then the Q-value of the reaction is nearly :

It has been estimated that the total power radiated by the sun is 3.8

1.	It has been estimated that the total power radiated by the sun is 3.8 xx 10^(26) J per second . The source of energy of stars is a thermonucleur fission reaction . Energy released in the process of fusion is due to mass defect . Q = Delta mc^(2) In a nuclear reaction , ""_(1)^(2) H + ""_(1)^(2) H to ""_(2)^(3) He + ""_(0)^(1)n . if the masses of ""_(1)^(2) H & ""_(2)^(3) He are 2.014741 amu and 3.016977 amu respectively . then the Q-value of the reaction is nearly :
Answer» 0.00352 Me v 3.63 Me v 0.082 MEV 2.45 Mev Solution :`""^(2) (""_(1) H^(2)) to ""_(2) He^(3) + ""_(1) n^(1) , Q = ?` `DELTA m = (2 xx 2.01474) - [3.016977 + 1.0086] = (4.029482 - 4.025577)` = `0.003904999` amu , `Q = Delta m xx 931.5 MeV = 3.6375 Me V = 3.63 MeV`