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Calculate the energy released in the fol...

Calculate the energy released in the following:
`._(1)H^(2) + ._(1)H^(3) rarr ._(2)He^(4) + ._(0)n^(1)`
(Given masses : `H^(2) = 2.014, H^(3) = 3.016, He = 4.003, n = 1.009 m_(u))`

A

16.76

B

26.38

C

13.26

D

23.275

Text Solution

Verified by Experts

The correct Answer is:
D
`Deltam=(2xx2.014-4.003) m_(u)`
`=(4.028-4.003)m_(u)`
Now ` 1 m_(u)=936` MeV
`therefore" "E=0.025xx936`
`=(25)/(100)xx936=(93.6)/(4)`
=23.4 MeV
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