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.