The minimum kinetic energy needed by an alpha particle to cause the nuclear reaction ¹⁶₇N + ⁴₂He → ¹₁H + ¹⁹₈O in a laboratory frame is n (in MeV). Assume that ¹⁶₇N is at rest in the laboratory frame. The masses of ¹⁶₇N, ⁴₂He, ¹₁H and ¹⁹₈O can be taken to be 16.006 u, 4.003, 1.008 u and 19.003 u, respectively, where 1 u = 930 MeV c⁻². The value of n is _________.

The correct answer is n = 2.33 MeV.

The nuclear reaction is:

¹⁶₇N + ⁴₂He → ¹₁H + ¹⁹₈O

Since ¹⁶₇N is the stationary target and ⁴₂He is the projectile, this is a threshold energy problem. We need to find the minimum kinetic energy T_th of the alpha particle in the lab frame. The given masses are:

M₁ = M(⁴₂He) = 4.003 u (projectile)
M₂ = M(¹⁶₇N) = 16.006 u (target, at rest)
M₃ = M(¹₁H) = 1.008 u
M₄ = M(¹⁹₈O) = 19.003 u
1 u = 930 MeV/c²

Step 1: Calculate the Q-value of the reaction.

Q = [M₁ + M₂ - M₃ - M₄] × c²

Q = [4.003 + 16.006 - 1.008 - 19.003] × 930 MeV

Q = [20.009 - 20.011] × 930 MeV

Q = (-0.002) × 930 MeV = -1.86 MeV

Since Q < 0, the reaction is endothermic — the alpha particle must supply a minimum threshold kinetic energy.

Step 2: Derive the threshold kinetic energy using the Lorentz invariant.

The square of the total 4-momentum is a Lorentz invariant. In the lab frame (target at rest), the invariant mass squared is:

s^{}={(E_1+M_2{c}^2)}^2-{p_{1}c}^2

Using E₁² = (p₁c)² + (M₁c²)² and E₁ = T + M₁c², this simplifies to:

s={\left(M_1{c}^2\right)}^2+{\left(M_2{c}^2\right)}^2+2M_2{c}^{2}T+2M_1{M}_2{c}^4

At the threshold condition, all product particles are produced at rest in the center-of-mass frame, so the minimum invariant mass is:

{\left(\sqrt{s}\right)}_{\min }=(M_3+M_4)c^2

Setting s_min = (M₃ + M₄)²c⁴ and solving for T_th:

T_{\mathrm{th}}=\frac{[{(M_3+M_4)}^2-{(M_1+M_2)}^2]c^2}{2M_2}

Step 3: Substitute values.

M₁ + M₂ = 4.003 + 16.006 = 20.009 u

M₃ + M₄ = 1.008 + 19.003 = 20.011 u

(M₃ + M₄)² = (20.011)² = 400.4401 u²

(M₁ + M₂)² = (20.009)² = 400.3601 u²

(M₃ + M₄)² - (M₁ + M₂)² = 400.4401 - 400.3601 = 0.08004 u²

Now substitute into the threshold formula (using 1 u × c² = 930 MeV):

T_{\mathrm{th}}=\frac{0.08004u\times c^2}{2\times 16.006}=\frac{0.08004\times 930\mathrm{MeV}}{32.012}

T_{\mathrm{th}}=\frac{74.44\mathrm{MeV}}{32.012}=2.325\mathrm{MeV}\approx 2.33\mathrm{MeV}

Verification using the Q-value form:

An equivalent and insightful way to express the threshold energy is:

T_{\mathrm{th}}=-Q\times \frac{M_1+M_2+M_3+M_4}{2M_2}

T_{\mathrm{th}}=1.86\times \frac{20.009+20.011}{2\times 16.006}=1.86\times \frac{40.02}{32.012}=1.86\times 1.2502\approx 2.33\mathrm{MeV}

Both methods confirm the result. The minimum kinetic energy needed by the alpha particle is n = 2.33 MeV.