Minimum kinetic energy needed for alpha particle nuclear reaction

To find the minimum kinetic energy (threshold energy, $K_{\text{th}}$) required by the $\alpha$-particle ($^4_2\text{He}$) to initiate the given nuclear reaction in the laboratory frame, we follow these steps:

Step 1: Calculate the $Q$-value of the nuclear reaction

The nuclear reaction is: $^{16}_7\text{N} + {}^4_2\text{He} \rightarrow {}^1_1\text{H} + {}^{19}_8\text{O}$

The given atomic masses are:

• Target nucleus ($^{16}_7\text{N}$): $m_T = 16.006\text{ u}$
• Projectile nucleus ($^4_2\text{He}$): $m_p = 4.003\text{ u}$
• Product nucleus ($^1_1\text{H}$): $m_H = 1.008\text{ u}$
• Product nucleus ($^{19}_8\text{O}$): $m_O = 19.003\text{ u}$

The mass change ($\Delta m$) for the reaction is given by: $\Delta m = (m_T + m_p) - (m_H + m_O)$ $\Delta m = (16.006 + 4.003)\text{ u} - (1.008 + 19.003)\text{ u}$ $\Delta m = 20.009\text{ u} - 20.011\text{ u} = -0.002\text{ u}$

Using $1\text{ u} = 930\text{ MeV}c^{-2}$, the $Q$-value of the reaction is: $Q = \Delta m \times 930\text{ MeV} = -0.002 \times 930\text{ MeV} = -1.86\text{ MeV}$

Since $Q < 0$, the reaction is endoergic (endothermic).

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Step 2: Calculate the threshold kinetic energy ($K_{\text{th}}$)

For a target nucleus at rest, conservation of linear momentum and energy requires the threshold kinetic energy of the projectile in the laboratory frame to be: $K_{\text{th}} = |Q| \left(1 + \frac{m_p}{m_T}\right)$

Substituting the known values: $K_{\text{th}} = 1.86 \times \left(1 + \frac{4.003}{16.006}\right)\text{ MeV}$ $K_{\text{th}} = 1.86 \times \left(\frac{20.009}{16.006}\right)\text{ MeV}$ $K_{\text{th}} = \frac{37.21674}{16.006}\text{ MeV} \approx 2.32517\text{ MeV}$

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Step 3: Final Answer

Rounding off to two decimal places, the value of $n$ is $2.33$ (or $2.325$).