Question

How much energy is released in the following reaction
7Li + p → α + α.
Atomic mass of 7Li = 7.0160 u and that of 4He = 4.0026 u.

Answer 1

Answer:

7Li+p→α+α

Atomic mass of 7Li - 7.0160 u and that of 4He - 4.0026 u.

Answer 2

Answer:

A::C

Solution :

Li7+p→α+α+E,

Li7=7.016u

α=4He=4.0026u,p=1.007276uE = Li^7 + P -2alpha=(7.016 + 1.007276)u-(2 xx 4.0026)u

=(8.023273−8.0052)u  

=0.018076u

=0.018076×931

=16.828=16.83MeV .

Hope this helps u