Answer to Question #108331 in Quantum Mechanics for Shivi

Question

A particle of mass M, initially at rest, decays into two particles with rest masses
m1
and
m2
respectively. Show that the total energy of the mass
m1
is:
M
c M m m
E
2
[ ]
2
2
2
1
2 2
1
 

Answer 1

$E_1+E_2=m_0c^2$

$\overrightarrow{p_1}+\overrightarrow{p_2}=0$

$E_1^2=m_1^2c^4$ and $p_1^2=2m_1E_1$

$(m_0c^2-E_1)^2=E_2^2$

$(m_0c^2-E_1)^2=$

$=(m_0^2c^4+m_1^2c^4-2m_0c^2m_1c^2)=$

$=c^4(m_0^2+m_1^2-2m_0m_1)=$

$=c^4(m_0^2+m_1^2)-c^22m_0E_1$

$E_2^2=m_2^2c^4$

$m_2^2c^4=c^4(m_0^2+m_1^2)-c^22m_0E_1\to m_2^2c^2=c^2(m_0^2+m_1^2)-2m_0E_1$

So, we have

$E_1=\frac{c^2}{2m_0}(m_0^2+m_1^2-m_2^2)$ Answer.