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Q.

Bombardment of lithium with protons gives rise to the following reaction :
 3Li7+1H12 2He4+Q. Find the Q– value of the reaction. The atomic masses of lithium, proton and helium are 7.016 u, 1.008 u and 4.004 u respectively.

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a

10.904 Mev

b

12.904 Mev

c

16.904 Mev

d

14.904 Mev

answer is B.

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Detailed Solution

Q=Δm×931MeV

=[m(Li)+m(H)-2m(He)]×931MeV

=[7.016+1.008-2×4.004]×931

=0.016×931=14.896MeV

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