A proton is bombarded on a stationary lithium nucleus. As a result of the collision, two `alpha`-particles are produced. If the direction of motion of the `alpha`-particles with the initial direction of motion makes an angles `cos^-1(1//4)`, find the kinetic energy of the striking proton. Given, binding energies per nucleon of `Li^7` and `He^4` are 5.60 and 7.06 MeV, respectively.
(Assume mass of proton `~~` mass of neutron).

Explanation

To solve the problem, we will follow these steps:

### Step 1: Calculate the Binding Energies
First, we need to calculate the total binding energy of the reactants and products.

- **Binding energy of reactants (Li-7)**:
- Lithium-7 has 7 nucleons, and the binding energy per nucleon is 5.60 MeV.
- Total binding energy of Li-7 = 7 nucleons × 5.60 MeV/nucleon = 39.2 MeV.

- **Binding energy of products (2 alpha particles)**:
- Each alpha particle (He-4) has 4 nucleons, and the binding energy per nucleon is 7.06 MeV.
- Total binding energy of 2 alpha particles = 2 × (4 nucleons × 7.06 MeV/nucleon) = 2 × 28.24 MeV = 56.48 MeV.

### Step 2: Calculate the Q-value
The Q-value of the reaction can be calculated using the difference in binding energies:

\[ Q = \text{Binding energy of products} - \text{Binding energy of reactants} \]
\[ Q = 56.48 \text{ MeV} - 39.2 \text{ MeV} = 17.28 \text{ MeV} \]

### Step 3: Apply Conservation of Momentum
In the collision, we can apply the conservation of momentum. The momentum before the collision (of the proton) must equal the total momentum after the collision (of the two alpha particles).

Let:
- \( K_p \) = Kinetic energy of the proton
- \( K_\alpha \) = Kinetic energy of each alpha particle
- \( \theta \) = angle between the direction of motion of the alpha particles and the initial direction of the proton, where \( \cos \theta = \frac{1}{4} \).

The momentum of the proton before the collision is given by:
\[ p_p = \sqrt{2 m_p K_p} \]

The momentum of the two alpha particles after the collision is:
\[ p_\alpha = 2 \sqrt{2 m_\alpha K_\alpha} \cos \theta \]

### Step 4: Equate Momenta
Since momentum is conserved:
\[ \sqrt{2 m_p K_p} = 2 \sqrt{2 m_\alpha K_\alpha} \cos \theta \]

### Step 5: Substitute Values
Assuming the mass of the proton \( m_p \) is approximately equal to the mass of the alpha particle \( m_\alpha \) (since an alpha particle consists of 2 protons and 2 neutrons, its mass is roughly 4 times that of a proton), we can simplify the equation.

Let \( m = m_p \approx m_\alpha/4 \):
\[ \sqrt{2 m K_p} = 2 \sqrt{2 m K_\alpha} \cdot \frac{1}{4} \]
Squaring both sides:
\[ 2 m K_p = \frac{1}{4} \cdot 4 \cdot 2 m K_\alpha \]
\[ K_p = K_\alpha \]

### Step 6: Relate Kinetic Energies
From the earlier calculation, we know:
\[ K_p = Q = 17.28 \text{ MeV} \]

### Final Answer
Thus, the kinetic energy of the striking proton is:
\[ \boxed{17.28 \text{ MeV}} \]

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