An alpha particle with kinetic energy 7.70 MeV collides with an ^14 N nucleus at rest, and the

An alpha particle with kinetic energy 7.70 MeV collides with...

An alpha particle with kinetic energy $7.70 \mathrm{MeV}$ collides with an ${ }^{14} \mathrm{~N}$ nucleus at rest, and the two transform into an ${ }^{17} \mathrm{O}$ nucleus and a proton. The proton is emitted at $90^{n}$ to the direction of the incident alpha particle and has a kinetic energy of $4.44 \mathrm{MeV}$. The masses of the various particles are alpha particle, $4.00260 \mathrm{u} ;{ }^{14} \mathrm{~N}$, $14.00307 \mathrm{u} ;$ proton, $1.007825 \mathrm{u} ;$ and ${ }^{17} \mathrm{O}, 16.99914 \mathrm{u} .$ In $\mathrm{MeV},$ what are (a) the kinetic energy of the oxygen nucleus and (b) the $Q$ of the reaction? (Hint: The speeds of the particles are much less than $c$ )

An alpha particle with kinetic energy $7.70

\mathrm{MeV}$ collides with an ${ }^{14}

\mathrm{~N}$ nucleus at rest, and the two

transform into an ${ }^{17} \mathrm{O}$

nucleus and a proton. The proton is emitted

at $90^{n}$ to the direction of the incident

alpha particle and has a kinetic energy of

$4.44 \mathrm{MeV}$. The masses of the

various particles are alpha particle, $4.00260

\mathrm{u} ;{ }^{14} \mathrm{~N}$,

$14.00307 \mathrm{u} ;$ proton, $1.007825

\mathrm{u} ;$ and ${ }^{17} \mathrm{O},

16.99914 \mathrm{u} .$ In $\mathrm{MeV},$

what
are (a) the kinetic energy of the oxygen

nucleus and (b) the $Q$ of the reaction?

(Hint: The speeds of the particles are much

less than $c$ )

Key Concepts

Conservation of Momentum

In nuclear reactions, conservation of momentum is essential for determining the velocities and directions of the reaction products. This principle states that the total linear momentum of a closed system remains constant if no external forces act on it, and it is particularly useful when analyzing collisions, especially when particles are emitted at different angles.

Conservation of Energy

Energy conservation in nuclear reactions involves accounting for all forms of energy before and after the reaction, including the kinetic energies of the particles and any additional energy due to mass defect or nuclear binding energy. Even when energy appears to be 'lost' or 'gained', it is simply converted into other forms, such as kinetic energy, and this balance is critical for solving for unknown variables in reaction kinematics.

Q Value of a Nuclear Reaction

The Q value represents the net energy change in a nuclear reaction. It is defined as the difference between the total rest mass energies of the initial particles and the final particles, indicating whether energy is released (exothermic reaction) or absorbed (endothermic reaction). This concept is central to understanding how mass differences in nuclear processes translate into kinetic energy or other forms of energy.

Non-relativistic Kinematics

When particle speeds are much less than the speed of light, as is often the case in many nuclear reactions, classical or non-relativistic kinematics can be applied. This simplifies the calculations by allowing the use of standard equations for momentum and kinetic energy without incorporating relativistic effects.

Mass-Energy Equivalence

Mass-energy equivalence, embodied in Einstein’s equation E=mc², underpins nuclear reactions by linking the mass defect (the difference in mass between reactants and products) to the energy released or absorbed in the reaction. This principle is crucial for accurately calculating the Q value, as even small differences in mass can correspond to significant amounts of energy.

Transcript

00:01 For this problem on the topic of relativity, we have an alpha particle with kinetic energy 7 .7 mega electron voles colliding with a nitrogen 14 nucleus which is at rest, and the two transform into an oxygen 17 nucleus and a proton.

00:13 The proton is emitted at 90 degrees to the direction of the incident alpha particle and has a kinetic energy of 4 .44 mega electron volts.

00:22 We are given the masses of the various particles and the alpha particle, and we want to find the kinetic energy of the oxygen nucleus, as well as the q value for the reaction.

00:32 Now along the x and y axes, momentum conservation gives us, firstly, m alpha, v alpha, equal to the mass of the oxygen molecule times the x component of its velocity, which means that we can write its x component of velocity to be the mass of the alpha particle m alpha times the mass of the oxygen molecule times v alpha, which is approximately 4 over 17 times v alpha.

01:09 For the y component, we have 0 is equal to the mass of the oxygen nucleus times the speed of the oxygen nucleus in the y direction, plus the mass of the proton times the speed of the proton.

01:26 This gives the y component of the velocity of the oxygen nucleus to be minus the mass of the proton over the mass of the oxygen times the speed of the proton.

01:41 And this is approximately minus 1 over 17 vp.

01:47 Now to complete the determinations, we need values for the initial speed of the alpha particle and the final speed of the proton.

01:54 And we can do this by rewriting the classical kinetic energy expression as k.

01:59 Is equal to a half m c squared beta squared and solve for beta and so for the proton we obtain a beta p to be the square root of two times the kinetic energy of the proton divided by mc squared for the proton which is the square root of two times four point four mega electron volts divided by 938 mega electron volts.

02:34 This gives us 0 .0973.

02:40 For the alpha particle, we have m -alpha c squared equal to its atomic mass 4 .0026 atomic mass units u times 931 .31 .5 mega -elelect electron volts per atomic mass unit, which gives us an energy of 3 ,728 mega electron volts.

03:09 Then we obtain beta alpha then to be the square root of 2k alpha divided by m alpha c squared, which is the square root of 2 times 7 .7...

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