The nuclear reaction `n+._5^(10)B rarr ._3^7Li + ._2^4He` is observed to occur even when very slow-moving neutrons `(M_ (n)=1.0087 am u)` strike a boron atom at rest. For a particular reaction in which `K_(n) =0` , the helium `(M_(He) =4.0026 am u)` is observed to have a speed of `9.30 xx 10^(6) m s^(-1)`. Determine (a) the kinetic energy of the lithium `(M_(Li) =7.0160 am u)` and (b) the `Q` value of the reaction.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Calculate the Kinetic Energy of Lithium

The kinetic energy (KE) of an object can be calculated using the formula:
\[ KE = \frac{1}{2} mv^2 \]
where:
- \( m \) is the mass of the object,
- \( v \) is the velocity of the object.

For lithium (\( \text{Li} \)):
- Mass \( M_{\text{Li}} = 7.0160 \, \text{amu} \)
- Velocity \( v = 9.30 \times 10^6 \, \text{m/s} \)

First, we need to convert the mass from atomic mass units (amu) to kilograms (kg). The conversion factor is:
\[ 1 \, \text{amu} = 1.660539 \times 10^{-27} \, \text{kg} \]
Thus, the mass of lithium in kg is:
\[ m_{\text{Li}} = 7.0160 \, \text{amu} \times 1.660539 \times 10^{-27} \, \text{kg/amu} = 1.165 \times 10^{-26} \, \text{kg} \]

Now we can calculate the kinetic energy:
\[ KE_{\text{Li}} = \frac{1}{2} (1.165 \times 10^{-26} \, \text{kg}) (9.30 \times 10^6 \, \text{m/s})^2 \]

Calculating \( v^2 \):
\[ (9.30 \times 10^6)^2 = 8.649 \times 10^{13} \, \text{m}^2/\text{s}^2 \]

Now substituting back into the kinetic energy formula:
\[ KE_{\text{Li}} = \frac{1}{2} (1.165 \times 10^{-26}) (8.649 \times 10^{13}) = 5.04 \times 10^{-13} \, \text{J} \]

### Step 2: Calculate the Q Value of the Reaction

The Q value of a nuclear reaction is given by the mass defect multiplied by \( c^2 \):
\[ Q = \Delta m \cdot c^2 \]
where \( c \) is the speed of light (\( c = 3 \times 10^8 \, \text{m/s} \)) and \( \Delta m \) is the difference in mass between the reactants and products.

The masses involved in the reaction are:
- Mass of neutron \( M_n = 1.0087 \, \text{amu} \)
- Mass of boron \( M_B = 10.0000 \, \text{amu} \)
- Mass of lithium \( M_{\text{Li}} = 7.0160 \, \text{amu} \)
- Mass of helium \( M_{\text{He}} = 4.0026 \, \text{amu} \)

The total mass of reactants:
\[ M_{\text{reactants}} = M_n + M_B = 1.0087 + 10.0000 = 11.0087 \, \text{amu} \]

The total mass of products:
\[ M_{\text{products}} = M_{\text{Li}} + M_{\text{He}} = 7.0160 + 4.0026 = 11.0186 \, \text{amu} \]

Now, calculate the mass defect:
\[ \Delta m = M_{\text{reactants}} - M_{\text{products}} = 11.0087 - 11.0186 = -0.0099 \, \text{amu} \]

Now, convert the mass defect to energy:
\[ Q = \Delta m \cdot 931.5 \, \text{MeV/amu} = -0.0099 \cdot 931.5 \, \text{MeV} = -9.23 \, \text{MeV} \]

To convert MeV to Joules, use:
\[ 1 \, \text{MeV} = 1.602 \times 10^{-13} \, \text{J} \]
Thus,
\[ Q = -9.23 \times 1.602 \times 10^{-13} \, \text{J} = -1.48 \times 10^{-12} \, \text{J} \]

### Final Answers:
(a) The kinetic energy of lithium is approximately \( 5.04 \times 10^{-13} \, \text{J} \).

(b) The Q value of the reaction is approximately \( -1.48 \times 10^{-12} \, \text{J} \).