The Q-value of a nuclear reaction and kinetic | vedclass

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Question

(D) Consider a nuclear reaction of the form $x + p \rightarrow y + b$,where $x$ is the target nucleus,$p$ is the projectile,$y$ is the product nucleus

Vedclass · Accepted Answer

$(K_{p} + Q) > 0$

Vedclass · Answer

(D) Consider a nuclear reaction of the form $x + p \rightarrow y + b$,where $x$ is the target nucleus,$p$ is the projectile,$y$ is the product nucleus,and $b$ is the emitted particle.The $Q$-value of a nuclear reaction is defined as the difference between the initial kinetic energy and the final kinetic energy,or equivalently,the energy released due to the mass defect: $Q = K_{f} + K_{b} - K_{p}$.Rearranging this equation,we get $Q + K_{p} = K_{f} + K_{b}$.Since the kinetic energies of the product particles ($K_{f}$ and $K_{b}$) must be non-negative (i.e.,$K_{f} \geq 0$ and $K_{b} \geq 0$),their sum must be greater than or equal to zero.Therefore,for the reaction to occur,the total energy available must satisfy the condition $(K_{p} + Q) > 0$.

The $Q$-value of a nuclear reaction and kinetic energy of the projectile particle,$K_{p}$,are related as

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