N-14 + alpha -> O-17 + proton

In summary, the interaction between N-14 and an alpha particle results in the production of O-17 and a high energy proton with a total energy of 13,083 MeV. The energy is shared between the two particles, with 6.5MeV being shared after the interaction. This is due to the conservation of energy, as the total energy before and after the interaction must remain constant. The mass of the alpha particle also plays a role in the energy distribution.
  • #1
Masafi
65
0
N-14 + alpha --> O-17 + proton

N-14 + alpha particle --> O-17 + high energy proton

alpha particle used was 7.70MeV.

Q1. Calculate the energy to be shared in the interaction.

Q2. Explain why the proton cannot take all of the available energy.



The following date is given for nuclide and their mass:
Hydrogen mass: 1.007825u, Helium 4.002603u, N-14 14.003074u, O-17 16.999132u



I tried converting the mass of N into energy in eV and adding it to 7.70MeV. Not sure what to do then.

The answer given to Q1 is 6.5MeV

The explanation for Q2 is not given.
 
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  • #2


Use conservation of energy: total energy initial = total energy final.
 
  • #3


vela said:
Use conservation of energy: total energy initial = total energy final.

Mass of N = 14.003074 * 1.66 x 10^(-27)
= 2.32 x 10^(-26) kg

Energy of N = 13075 MeV

Initial E = 13075 + 7.70 = 13,083 MeV


Final E from the Rest Mass of O and the proton = 16,814 MeV

So change in E = 3731MeV

and this is for the inital E, because that is the one which has less E when just taking into account the rest mass.

The answer should be 6.5MeV which is shared AFTER the interaction?
 
  • #4


What about the mass of the alpha particle?
 

1. What is the significance of the reaction N-14 + alpha -> O-17 + proton?

This reaction is known as a nuclear fusion reaction, where two atomic nuclei combine to form a new nucleus. It is an important process in the formation of new elements and the release of energy, and is also responsible for powering stars and nuclear reactors.

2. How does the reaction N-14 + alpha -> O-17 + proton occur?

This reaction occurs when a nitrogen-14 atom (containing 14 protons and 14 neutrons) collides with an alpha particle (containing 2 protons and 2 neutrons). The impact of the collision causes the nuclei to fuse together, forming oxygen-17 (with 17 protons and 14 neutrons) and releasing a proton.

3. What is the role of N-14 and alpha particles in this reaction?

The nitrogen-14 atom is the reactant in this reaction, meaning it is the starting material that undergoes a change. The alpha particle, which is a helium nucleus, acts as a "projectile" that collides with the nitrogen-14 atom to initiate the reaction.

4. What is the energy released in the reaction N-14 + alpha -> O-17 + proton?

The energy released in this reaction is approximately 4.15 MeV (million electron volts). This energy is released in the form of kinetic energy of the products (oxygen-17 and the proton) and also in the form of gamma rays, which are high-energy electromagnetic radiation.

5. What are the applications of the N-14 + alpha -> O-17 + proton reaction?

This reaction has many important applications, including in nuclear power generation, where it is used to produce energy in nuclear reactors. It is also used in nuclear medicine, as it is the basis of positron emission tomography (PET) scans. Additionally, this reaction plays a crucial role in the formation of elements in stars and in the universe as a whole.

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