`|Q|` value of the reaction, `N^(4)+alpha rarrO^(17)+p`
[the masses of `N^(14),He^(4),p,O^(17)` are respectively `14.00374 u, 4.0026 u,1.00783 u` and `16.99913 u` is `x//5MeV`. Value of `x` is nearly `[1 u=931.5 MeV]`

Explanation

To find the Q value of the nuclear reaction \( N^{14} + \alpha \rightarrow O^{17} + p \), we follow these steps:

### Step 1: Write down the masses involved in the reaction
- Mass of \( N^{14} \) (Nitrogen): \( 14.00374 \, u \)
- Mass of \( \alpha \) (Helium): \( 4.0026 \, u \)
- Mass of \( O^{17} \) (Oxygen): \( 16.99913 \, u \)
- Mass of \( p \) (Proton): \( 1.00783 \, u \)

### Step 2: Calculate the total mass of the reactants
The total mass of the reactants is the sum of the masses of nitrogen and the alpha particle:
\[ \text{Total mass of reactants} = m(N^{14}) + m(\alpha) = 14.00374 \, u + 4.0026 \, u = 18.00634 \, u \]

### Step 3: Calculate the total mass of the products
The total mass of the products is the sum of the masses of oxygen and the proton:
\[ \text{Total mass of products} = m(O^{17}) + m(p) = 16.99913 \, u + 1.00783 \, u = 18.00696 \, u \]

### Step 4: Calculate the mass defect
The mass defect \( \Delta m \) is the difference between the total mass of the reactants and the total mass of the products:
\[ \Delta m = \text{Total mass of reactants} - \text{Total mass of products} \]
\[ \Delta m = 18.00634 \, u - 18.00696 \, u = -0.00062 \, u \]

### Step 5: Calculate the Q value
The Q value is related to the mass defect by the formula:
\[ Q = \Delta m \times 931.5 \, \text{MeV/u} \]
Substituting the value of \( \Delta m \):
\[ Q = -0.00062 \, u \times 931.5 \, \text{MeV/u} \approx -0.577 \, \text{MeV} \]

### Step 6: Relate Q value to the given expression
According to the problem, the Q value is given as \( \frac{x}{5} \, \text{MeV} \). Therefore, we set:
\[ \frac{x}{5} = -0.577 \Rightarrow x = -0.577 \times 5 \approx -2.885 \]

### Step 7: Rounding and final answer
Since the value of \( x \) is typically expressed as a positive integer in such contexts, we take the absolute value and round it appropriately. Thus, the final answer for \( x \) is approximately \( 3 \).

### Final Answer
The value of \( x \) is \( 3 \).

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