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Calculate the energy of the following nu...

Calculate the energy of the following nuclear reaction:
`._(1)H^(2)+._(1)H^(3) to ._(2)He^(4) + ._(0)n^(1)+Q`
Given: `m(._(1)H^(2))=2.014102u, m(._(1)H^(3))=3.016049u, m(._(2)He^(4))=4.002603u, m(._(0)n^(1))=1.008665u`

A

`-4.03` MeV

B

`-2.01` MeV

C

2.01 MeV

D

4.03 MeV

Text Solution

Verified by Experts

The correct Answer is:
A
`""_(1)H^(1)H+""_(1)^(3)H rarr""_(1)^(2)H+""_(1)^(2)H`
The Q - value of the above reaction is
`Q=DeltaMc^(2)[1.007825+3.016049-2xx2.014102]uxxc^(2)`
`=-4.33xx10^(-3)xxc^(2)`
`because 1u=931 MeV/c"^(2)`
`therefore" "Q=-4.33xx10^(-3)xx"931 MeV "=-"4.03 MeV"`
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