The $$Q$$-value of a nuclear reaction and kinetic energy of the projectile particle, $$K_p$$ are related as

We need to find the relationship between the Q-value of a nuclear reaction and the kinetic energy of the projectile particle $$K_p$$.

The Q-value of a nuclear reaction is defined as:

$$Q = (\text{Total K.E. of products}) - (\text{Total K.E. of reactants})$$

Or equivalently: $$Q = (\text{Sum of rest masses of reactants} - \text{Sum of rest masses of products}) \times c^2$$

For a nuclear reaction to proceed, the total kinetic energy of the products must be positive (or zero at threshold):

$$\text{Total K.E. of products} \geq 0$$

Since $$Q = K_{\text{products}} - K_{\text{reactants}}$$, it follows that $$K_{\text{products}} = Q + K_{\text{reactants}}$$. Because the target is typically at rest, $$K_{\text{reactants}} = K_p$$, so $$K_{\text{products}} = K_p + Q$$.

Requiring the total kinetic energy of the products to be positive gives $$K_p + Q > 0$$. Strictly, $$K_p + Q \geq 0$$, but in the center-of-mass frame some kinetic energy is always retained, so in the lab frame we require $$K_p + Q > 0$$ for the reaction to occur and produce observable products.

The correct answer is Option D.