Question

Consider the nuclear rxnx in which alpha particle collides ith stationary nitrogen to givw O17 and Pproton. incomng alpha aprticle has KE=4MeV and outgoing proton makes 6- deg with incoming direction and has ke=2.09MeV. find Q value

• A. -1.52 MeV
• B. -2.00 MeV
• C. -1.00 MeV
• D. -0.50 MeV

Answer 1

Answer: (A)

-1.52 MeV

Answer 2

Solution:

We are given the reaction

$$\alpha(4\text{ u}) + \,^{14}\!N(14\text{ u}) \to \,^{17}\!O(17\text{ u}) + p(1\text{ u})$$

with the following data (all energies in MeV):

• Incident alpha has kinetic energy $K_\alpha = 4$.
• The proton (mass 1 u) is emitted at an angle $6^\circ$ with kinetic energy $K_p = 2.09$.
• The nitrogen target is at rest.
• Let the oxygen recoil’s kinetic energy be $K_O$.

Step 1. Energy Conservation

The energy balance in the lab frame is

$$K_\alpha + Q = K_p + K_O \quad \implies \quad K_O = 4 + Q - 2.09 = Q + 1.91.$$

Step 2. Momentum Conservation

In the lab frame, momentum is conserved in both $x$ (initial beam direction) and $y$ directions.

Define the momenta by:

$$p_i = \sqrt{2m_i K_i}.$$

• For the alpha: $$p_\alpha = \sqrt{2 \cdot 4 \cdot 4} = \sqrt{32} \approx 5.657.$$
• For the proton: $$p_p = \sqrt{2 \cdot 1 \cdot 2.09} = \sqrt{4.18} \approx 2.044.$$
• For the oxygen recoil: $$p_O = \sqrt{2 \cdot 17 \cdot (Q+1.91)} = \sqrt{34(Q+1.91)}.$$

Momentum components:

1. Y-direction:

Since the initial momentum in $y$ is zero,

$$p_p \sin6^\circ = p_O \sin\theta_O.$$

Thus,

$$\sin\theta_O = \frac{p_p\sin6^\circ}{p_O}.$$

Using $\sin6^\circ \approx 0.1045$,

$$p_p\sin6^\circ \approx 2.044 \times 0.1045 \approx 0.2137.$$

So,

$$\sin\theta_O = \frac{0.2137}{\sqrt{34(Q+1.91)}}.$$

2. X-direction:

$$p_\alpha = p_p \cos6^\circ + p_O\cos\theta_O.$$

With $\cos6^\circ \approx 0.9945$,

$$2.044 \cos6^\circ \approx 2.044 \times 0.9945 \approx 2.034.$$

Thus,

$$5.657 = 2.034 + \sqrt{34(Q+1.91)}\cos\theta_O.$$

Also, note that

$$\cos\theta_O = \sqrt{1-\sin^2\theta_O} = \sqrt{1 - \frac{(0.2137)^2}{34(Q+1.91)}}.$$

Let

$$X = 34(Q+1.91).$$

Then the $x$-component equation becomes

$$5.657 = 2.034 + \sqrt{X} \sqrt{1 - \frac{0.0457}{X}},$$

since $0.2137^2 \approx 0.0457$. Rearranging,

$$3.623 = \sqrt{X}\sqrt{1 - \frac{0.0457}{X}}.$$

Squaring both sides:

$$(3.623)^2 = X\left(1-\frac{0.0457}{X}\right) = X-0.0457.$$

So,

$$X = (3.623)^2 + 0.0457 \approx 13.128 + 0.0457 \approx 13.1737.$$

Recall $X = 34(Q+1.91)$ hence,

$$34(Q+1.91)= 13.1737 \quad \implies \quad Q+1.91 = \frac{13.1737}{34} \approx 0.38746.$$

Thus,

$$Q \approx 0.38746 - 1.91 \approx -1.52254\,\text{MeV}.$$

Final Answer:

$$Q \approx -1.52\,\text{MeV}$$