Question

1. A Lithium-5 atom at rest decays into a proton and an a particle with the release of 3.15*10-13 J of kinetic energies total of the proton and the a particle. Determine the velocities of the proton and the a particle after the decay. You can assume the masses are ma = 4mp = 6.64*10-27 kg.

Answer 1

Initially the lithium atom is at rest. So the momentum is 0.

Now after the decay let the velocities of proton and alpha particles are $v_p, v_{\alpha}$

So the momentum of the system is,

$P=m_pv_p+m_{\alpha}v_{\alpha}$

Now from momentum conservation, the net momentum after the decay is zero.

$m_pv_p+m_{\alpha}v_{\alpha}=0$

тр"р -та'а

тр"р - 4mpo

$v_p=-4v_{\alpha}$

Now the total kinetic energy is,

$KE=\frac{1}{2}m_pv_p^2+\frac{1}{2}m_{\alpha}v_{\alpha}^2$

$KE=\frac{1}{2}m_p(-4v_{\alpha})^2+\frac{1}{2}(4m_p)v_{\alpha}^2$

$KE=8m_pv_{\alpha}^2+2m_pv_{\alpha}^2=10m_pv_{\alpha}^2$

Now,

$10m_pv_{\alpha}^2=3.15\times10^{-13}~J$

$10\times\frac{6.64\times10^{-27}}{4}v_{\alpha}^2=3.15\times10^{-13}$

$v_{\alpha}^2=3.15\times10^{-13}\times\frac{4}{6.64\times10^{-27}\times10}$

$v_{\alpha}^2=1.9\times10^{13}$

$v_{\alpha}=4.36\times10^{6}~m/s$

So velocity of the alpha particle is $v_{\alpha}=4.36\times10^{6}~m/s$

Velocity of the proton is $v_p=4v_{\alpha}=1.74\times10^{7}~m/s$