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It has been estimated that the total pow...

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 :

A

0.00352 Me v

B

3.63 Me v

C

0.082 Mev

D

2.45 Mev

Text Solution

Verified by Experts

The correct Answer is:
B
`""^(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`
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