Bombardment of lithium with protons gives rise to the following reaction :
`._3Li^7 + ._1H^1 to 2 (._2He^4)+Q`. Find the Q-value of the reaction. The atomic masses of lithium, proton and helium are 7.016 u, 1.008 u and 4.004 u respectively.

Options

A. 12.904 Mev
B. 14.904 Mev
C. 10.904 Mev
D. 16.904 Mev

Answer

To find the Q-value of the reaction \( _3Li^7 + _1H^1 \to 2 \, _2He^4 + Q \), we will follow these steps:

### Step 1: Write down the atomic masses
- Mass of Lithium (\( _3Li^7 \)): \( 7.016 \, u \)
- Mass of Proton (\( _1H^1 \)): \( 1.008 \, u \)
- Mass of Helium (\( _2He^4 \)): \( 4.004 \, u \)

### Step 2: Calculate the total mass of the reactants
The total mass of the reactants is the sum of the masses of lithium and the proton:
\[ \text{Total mass of reactants} = \text{Mass of Lithium} + \text{Mass of Proton} \]
\[ = 7.016 \, u + 1.008 \, u = 8.024 \, u \]

### Step 3: Calculate the total mass of the products
Since there are 2 helium nuclei produced, the total mass of the products is:
\[ \text{Total mass of products} = 2 \times \text{Mass of Helium} \]
\[ = 2 \times 4.004 \, u = 8.008 \, u \]

### Step 4: Calculate the mass defect (\( \Delta m \))
The mass defect (\( \Delta m \)) is given by the difference between the total mass of the reactants and the total mass of the products:
\[ \Delta m = \text{Total mass of reactants} - \text{Total mass of products} \]
\[ = 8.024 \, u - 8.008 \, u = 0.016 \, u \]

### Step 5: Convert mass defect to energy (Q-value)
The Q-value can be calculated using the formula:
\[ Q = \Delta m \times 931 \, \text{MeV/u} \]
Substituting the value of \( \Delta m \):
\[ Q = 0.016 \, u \times 931 \, \text{MeV/u} = 14.896 \, \text{MeV} \]

### Final Answer
Thus, the Q-value of the reaction is approximately:
\[ Q \approx 14.896 \, \text{MeV} \]