1.

Bombardment of lithium with protons gives rise to the following reaction : `._(3)^(7)Li + ._(1)^(1)H rarr 2[._(2)^(4)He]+Q`. Find the Q - value of the reaction. The atomi mases of lithium, proton and helium are 7.016u, 1.0084 and 4.004 u respectively.

Answer» Mass of Lithrium = 7.016 u
`m_(P)=1.008 u`
Mass of Helium = 4.004 u,
`u=931.5 MeV`
Q = [Total mass of the reactants - Total mass of the products `c^(2)`
= [Mass of Lithium `+ m_(P) -` (2 `xx` mass of Helium)] `xx 931.5 MeV`
`= [7.016+1.008-2(4.004)]xx931.5 MeV`
`=[8.024-8.008]xx931.5 MeV`
`therefore` Energy `Q=0.016xx931.5`
`= 14.904 MeV`


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