As,system has initial momentum is zero and no external interaction there,Momentum of the system is conserved.
Therefore, momentum
"p_{os}+p_{\\gamma}=0\n\n\n\u200b"
We have given that, emitted photon has energy is "8.6keV"
Since,
"E_{\\gamma}=p_{\\gamma}c\\\\ \\implies p_{\\gamma}=\\frac{E_{\\gamma}}{c}=8.6keV\/c"Thus,
"p_{os}=-8.6keV\/c"
Kinetic energy is
"K=\\frac{p^2}{2m}"
Hence,
"K_{os}=2.088\\times 10^{-4}eV"
Comments
Leave a comment