The minimum kinetic energy needed by an alpha particle to cause the nuclear reaction $$^{16}_7N + ^4_2He \to ^1_1H + ^{19}_8O$$ in a laboratory frame is $$n$$ (in MeV). Assume that $$^{16}_7N$$ is at rest in the laboratory frame. The masses of $$^{16}_7N$$, $$^4_2He$$, $$^1_1H$$ and $$^{19}_8O$$ can be taken to be 16.006 $$u$$, 4.003 $$u$$, 1.008 $$u$$ and 19.003 $$u$$, respectively, where 1 $$u$$ = 930 MeV$$c^{-2}$$. The value of $$n$$ is ______.

The reaction is a two-body reaction of the form $$a + A \rightarrow b + B$$ where the projectile $$a$$ is the $$\alpha$$-particle, the target $$A$$ is the stationary $$^{16}_7N$$ nucleus and the products are the proton $$b$$ and the $$^{19}_8O$$ nucleus $$B$$.

Step 1 : Calculate the Q-value of the reaction.
Initial rest-mass $$m_i = m_{\alpha} + m_{N} = 4.003\,u + 16.006\,u = 20.009\,u$$
Final rest-mass $$m_f = m_{p} + m_{O} = 1.008\,u + 19.003\,u = 20.011\,u$$
Mass difference $$\Delta m = m_i - m_f = 20.009\,u - 20.011\,u = -0.002\,u$$
Since $$1\,u = 930\;{\rm MeV}\,c^{-2}$$, the Q-value is $$Q = \Delta m\,c^{2} = (-0.002\,u)\times 930\;{\rm MeV} = -1.86\;{\rm MeV}$$ Thus the reaction is endothermic and needs at least $$1.86\;{\rm MeV}$$ of energy just to balance mass.

Step 2 : Threshold energy in the laboratory frame.
Because the target nucleus is at rest, conservation of momentum requires some of the projectile’s kinetic energy to appear as kinetic energy of the products. For a two-body reaction with target at rest, the minimum (threshold) kinetic energy $$T_{\text{th}}$$ of the projectile is

$$T_{\text{th}} = (-Q)\left(1 + \frac{m_a}{m_A}\right) \quad -(1)$$
where $$m_a$$ is the projectile mass and $$m_A$$ the target mass (both in the same units).

Here $$m_a = m_{\alpha} = 4.003\,u,\qquad m_A = m_{N} = 16.006\,u$$
so $$\frac{m_a}{m_A} = \frac{4.003}{16.006} \approx 0.2501$$

Insert the values into $$(1)$$: $$T_{\text{th}} = 1.86\,{\rm MeV}\;\Bigl(1 + 0.2501\Bigr) = 1.86\,{\rm MeV}\times 1.2501 \approx 2.325\;{\rm MeV}$$

Step 3 : Final answer.
The minimum kinetic energy required by the $$\alpha$$-particle is therefore approximately $$2.32\;{\rm MeV}$$.

Answer: 2.32-2.33 MeV