Angle between two alpha particle tracks due to proton bombardment of a nuclei

[Antoha1]

Homework Statement

A lithium nucleus is bombarded with a proton with kinetic energy E and mass m. This interaction produces two alpha particles of equal energy. At what angle do the alpha particles
escape from each other? The mass of the alpha particle is 𝑚𝑎, and that of the lithium nucleus is 𝑚𝑙𝑖.

Relevant Equations

$E_{rest}=mc^2$; $E=\frac{1}{2}mv^{2}$; $p=mv$

I apply energy conservation law:
$$E+(m_{li}+m)c^2=2m_{\alpha}(c^2+\frac{1}{2}v_{\alpha}^2)~~~~~~~~(1)$$
where $v_{\alpha}$ is speed of alpha particle.
then I apply the law of conservation of momentum:
$$mv=2m_{\alpha}v_{\alpha}cos\frac{\theta}{2}~~~~~(2)$$
Where $v$ is proton's speed before colision and $\theta$ is angle between alpha particles.
Because $E=\frac{1}{2}mv^{2} \Longrightarrow v=\sqrt{\frac{2E}{m}}$
Then $cos\frac{\theta}{2}$ from (2):
$$cos\frac{\theta}{2}=\frac{\sqrt{2mE}}{2m_{\alpha}v_{\alpha}}~~~~~~(3)$$
$v_{\alpha}$ from (1):
$$v_{\alpha}=\sqrt{\frac{E+(m_{li}+m-2m_{\alpha})c^2}{m_{\alpha}}}$$
After putting alpha particle's speed into 3rd equation, angle between particles is:
$$\theta=2arccos\sqrt{\frac{mE}{2m_{\alpha}\left[ E+(m_{li}+m-2m_{\alpha})c^2 \right]}}$$

Is my solution correct?

 

[kuruman]

I don't think it is correct to write the momentum of the proton as $mv$ because that is not the relativistic expression. You are given that the total relativistic energy of the proton is $E$. The momentum of the proton $p$ is what you get when you solve the expression $$E=\sqrt{p^2c^2+m_p^2c^4}$$ for $p$ in terms of the proton's rest mass $m_p$ and the given energy $E.$

Assuming that the alpha particles have the same energy, you should likewise use the total relativistic energy to write the energy conservation equation as $$E=2\sqrt{p_{\alpha}^2c^2+m_{\alpha}^2c^4}$$ and then solve for $p_{\alpha}$ in terms of $E$ to use in the momentum conservation equation.

 

[Antoha1]

So basically, mass of lithium given is useless in this problem?

 

[kuruman]

So basically, mass of lithium given is useless in this problem?

Oops, I carelessly made it look that it is useless. The second equation should be $$E+m_{Li}c^2=2\sqrt{p_{\alpha}^2c^2+m_{\alpha}^2c^4}.$$ Sorry about that.

 

[Antoha1]

ok, so adding rest energy of lithium, but what about rest energy of proton, because I am thinking that only its kinetic E is given according to statement

 

[PeroK]

ok, so adding rest energy of lithium, but what about rest energy of proton, because I am thinking that only its kinetic E is given according to statement

First, for relativistic collisions it is far better to work using mass, energy and momentum. The key equations involve these quantities.

Forget about velocity for now. If you need the velocity, you can get it from the momentum at the end.

Second, total energy and kinetic energy are related by $E_{tot} = E_{kin} + mc^2$. It doesn't really matter whether you are given total energy or kinetic energy. Solve the problem using total energy and you can always substitute in kinetic energy and mass for total energy at the end.

Note that normally $E$ is used for total energy and $T$ for kinetic energy. If the question gives $E$ as the kinetic energy, that is a bit naughty, IMHO. It means we need a new symbol for total energy. I find this a bit annoying.

Third, this allows you to treat these problems somewhat formulaically.

We have an incident particle with total energy, magnitiude of momentum and mass $E_1, p_1, m_1$. We have a target particle with total energy, momentum and mass $m_2c^2, 0, m_2$.

And, we have two alpha particles with energy, momentum and mass $E_3, p_3, m_3$, separated by an angle $\theta$.

We can write down conservation of energy:
$$E_1 + m_2c^2 = 2E_3$$And, conservation of momentum (in the original direction):
$$p_1 = 2p_3\cos(\frac \theta 2)$$And, for each particle we have:
$$E_n^2 = p_n^2c^2 + m_n^2c^4$$All we have to do is eliminate $p_3$ from the momentum equation. If we square that, then we can replace $p_3$ with $E_3$ and $m_3$. We know $m_3$ and, we can eliminate $E_3$ using the first equation.

Also, note that $p_1$ can be replaced by $E_1$ and $m_1$, which we know.

Think general: energy, momentum and the energy-momentum equation for each particle. Square, substitute and eliminate. And you're done.

 

[kuruman]

Note that normally $E$ is used for total energy and $T$ for kinetic energy. If the question gives $E$ as the kinetic energy, that is a bit naughty, IMHO.

IMHO 2.

 

[Antoha1]

my other question is: if particle's speed has been increased lets say from 0 to around c, the mass of it increases, lets say rest mass is m and relativistic (with speed) M. Lets say that the particle got energy from electric field E=qV. So conservation of energy would be $mc^2+qV=Mc^2+K$ K is kinetic energy and q is charge.

 

[PeroK]

my other question is: if particle's speed has been increased lets say from 0 to around c, the mass of it increases, lets say rest mass is m and relativistic (with speed) M.

Unless you are using an old textbook, the rest mass $m$ is the mass in modern physics. Relativistic mass is no longer used. In any case:

The rest mass energy is $mc^2$.

The total energy is $E = \gamma mc^2$

The kinetic energy is $T = E - mc^2 = (\gamma - 1)mc^2$.