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Question

A nucleus 220X at rest decays emitting an αparticle. If energy of daughter nucleus is 0.2MeV,Q value of the reaction is

A
10.8MeV
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B
10.9MeV
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C
11MeV
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D
11.1MeV
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Solution

The correct option is B 11MeV
zX220z2Y216+2He4
From energy and momentum conservation we can derive the equation for K.E. of alpha particle as K.E. of α particle=[(A4)/4]×K.E. of daughter nucles
=[(2204)/4]×0.2MeV=10.8MeV
value of the reaction is given by Q=K.E. of alpha particle +K .E of daughter nucleus
=10.8MeV+0.2MeV=11MeV

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