A proton, within a nucleus decays into a neutron during a process called β+ decay which occurs is some unstable nuclei. This process is represented by the equation ZAX→Z− 1AY+e++VeLet m( ZAX) represent the mass of an atom of ZAX, and c represent the velocity of light in vaccum. Assume that ve is massless. The Q value of the process is given by

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[m(ZAX)−m(z−1AY)−2me+]c2

[m( ZAX)−m(z−1AY)]c2

[m(ZAX)−m(z−1AY)−me+]c2

[m(ZAX)−m(z−1AY)+me+]c2

answer is B.

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Detailed Solution

Q=c2[mnum( ZAX)−mnum(Z−1AY)−me+] By definition.Putting,mnuc( ZAX)=m( ZAX)−Zme+and similarly for Z−1AY,we get the final result.[m(ZAX)−m(z−1AY)]c2

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