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Calculate the energy released in joules ...

Calculate the energy released in joules and `MeV` in the following nuclear reaction:
`._(1)H^(2) + ._(1)H^(2) rarr ._(2)He^(3) + ._(0)n^(1)`
Assume that the masses of `._(1)H^(2)`, `._(2)He^(3)`, and neutron `(n)`, respectively, are 2.40, 3.0160, and 1.0087 in amu.

Text Solution

Verified by Experts

`Delta m [2 xx 2.0141] - 3.0160 - 1.0087 = 3.5 xx 10^(-3) "amu"`
`:. Delta E = Delta m xx 931.478 MeV`
`= 3.5 xx 10^(-3) xx 931.478 = 3.260 MeV`
`:. Delta E = 5.223 xx 10^(-13) J`
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