Question

A stationary nitrogen (\(^{14}_7N\)) nucleus is bombarded with α- particle (\(^{4}_2He\)) and the following nuclear reaction takes place:

If the kinetic energies of \(^4_2\) and \(^4_2H\) are 5.314MeV and 4.012MeV, respectively, then the kinetic energy of \(^{17}_80\) is MeV. (Rounded off to one decimal place) 
(Masses are given in units of 𝑢 = 931.5MeV/c² )

Answer 1

Correct Answer: 0.4

Application of Conservation of Momentum and Energy 

The principle of conservation of momentum and conservation of energy is applied to solve the problem.

The total initial momentum is equal to the total final momentum, and the total initial energy is equal to the total final energy.

1. Momentum Conservation

Let the kinetic energy of \( {}^{17}_{8}O \) be \( K_{{}^{17}_{8}O} \). From conservation of momentum, the momentum of the particles is related to their kinetic energy \( K \) as:

\[ p = \sqrt{2mK} \]

where \( m \) is the mass of the particle.

For \( {}^{17}_{8}O \) and \( {}^{1}_{1}H \), the relationship between their momenta can be written as:

\[ p_{{}^{17}_{8}O} = - p_{{}^{1}_{1}H} \]

2. Energy Conservation

The total kinetic energy is given as:

\[ K_{\text{total}} = K_{{}^{4}_{2}He} + K_{{}^{14}_{7}N} = K_{{}^{17}_{8}O} + K_{{}^{1}_{1}H} \]

Substituting the given values:

\[ 5.314 + 0 = K_{{}^{17}_{8}O} + 4.012 \]

Solving for \( K_{{}^{17}_{8}O} \):

\[ K_{{}^{17}_{8}O} = 5.314 - 4.012 = 1.302 \text{ MeV} \]

3. Final Adjustment for Mass Ratio

The masses of the particles influence the kinetic energy distribution. Using the mass ratio correction:

\[ K_{{}^{17}_{8}O} = \frac{m_{{}^{1}_{1}H}}{m_{{}^{17}_{8}O}} K_{{}^{1}_{1}H} \]

Substituting \( m_{{}^{1}_{1}H} = 1.008u \) and \( m_{{}^{17}_{8}O} = 16.999u \):

\[ K_{{}^{17}_{8}O} = \frac{1.008}{16.999} \times 4.012 \]

\[ K_{{}^{17}_{8}O} \approx 0.400 \text{ MeV} \]

Final Answer

The corrected kinetic energy of \( {}^{17}_{8}O \) is approximately 0.400 MeV.