To solve the problem step by step, we will follow the outlined approach to find the kinetic energy of lithium and the Q value of the reaction.
### Given Data:
- Mass of neutron, \( M_n = 1.0087 \, \text{amu} \)
- Mass of boron, \( M_B = 10 \, \text{amu} \) (assumed for \( _5^{10}B \))
- Mass of lithium, \( M_{Li} = 7.0160 \, \text{amu} \)
- Mass of helium, \( M_{He} = 4.0026 \, \text{amu} \)
- Speed of helium, \( v_{He} = 9.30 \times 10^6 \, \text{m/s} \)
- Kinetic energy of neutron, \( K_n = 0 \)
### Step 1: Calculate the Kinetic Energy of Lithium
1. **Conservation of Momentum**:
Since the neutron is at rest and has no kinetic energy, the initial momentum of the system is zero. Therefore, the final momentum must also be zero:
\[
0 = p_{Li} + p_{He}
\]
Where:
\[
p_{Li} = M_{Li} \cdot v_{Li}
\]
\[
p_{He} = M_{He} \cdot v_{He}
\]
Thus, we can express the momentum of lithium as:
\[
M_{Li} \cdot v_{Li} = -M_{He} \cdot v_{He}
\]
From this, we can find the velocity of lithium:
\[
v_{Li} = -\frac{M_{He} \cdot v_{He}}{M_{Li}}
\]
2. **Calculate Kinetic Energy of Lithium**:
The kinetic energy of lithium is given by:
\[
KE_{Li} = \frac{1}{2} M_{Li} v_{Li}^2
\]
Substituting \( v_{Li} \):
\[
KE_{Li} = \frac{1}{2} M_{Li} \left(-\frac{M_{He} \cdot v_{He}}{M_{Li}}\right)^2
\]
Simplifying:
\[
KE_{Li} = \frac{1}{2} \frac{M_{He}^2 \cdot v_{He}^2}{M_{Li}}
\]
3. **Substituting Values**:
Convert the masses from amu to kg using \( 1 \, \text{amu} = 1.66 \times 10^{-27} \, \text{kg} \):
\[
M_{He} = 4.0026 \, \text{amu} \times 1.66 \times 10^{-27} \, \text{kg/amu} = 6.645 \times 10^{-27} \, \text{kg}
\]
\[
M_{Li} = 7.0160 \, \text{amu} \times 1.66 \times 10^{-27} \, \text{kg/amu} = 1.165 \times 10^{-26} \, \text{kg}
\]
Now substituting the values:
\[
KE_{Li} = \frac{1}{2} \cdot \frac{(6.645 \times 10^{-27})^2 \cdot (9.30 \times 10^6)^2}{1.165 \times 10^{-26}}
\]
4. **Calculating**:
After calculating, we find:
\[
KE_{Li} \approx 1.02 \, \text{MeV}
\]
### Step 2: Calculate the Q Value of the Reaction
1. **Calculate Kinetic Energy of Helium**:
\[
KE_{He} = \frac{1}{2} M_{He} v_{He}^2
\]
Substituting values:
\[
KE_{He} = \frac{1}{2} \cdot 4.0026 \cdot 1.66 \times 10^{-27} \cdot (9.30 \times 10^6)^2
\]
After calculation, we find:
\[
KE_{He} \approx 1.80 \, \text{MeV}
\]
2. **Calculate Q Value**:
The Q value is given by:
\[
Q = KE_{He} + KE_{Li} - KE_{reactants}
\]
Since the initial kinetic energy of the reactants is zero:
\[
Q = KE_{He} + KE_{Li}
\]
Substituting the values:
\[
Q = 1.80 \, \text{MeV} + 1.02 \, \text{MeV} = 2.82 \, \text{MeV}
\]
### Final Answers:
(a) Kinetic Energy of Lithium: \( KE_{Li} \approx 1.02 \, \text{MeV} \)
(b) Q Value of the Reaction: \( Q \approx 2.82 \, \text{MeV} \)
