Since, system has initial momentum is zero and no external interaction is going on with the system,Momentum of the system is conserved, hence, by conservation law of momentum
"p_{osmium}+p_{\\gamma}=0"We have given that, emitted photon has energy is
"8.6keV=8.6\\times 10^3eV"Thus,
"E_{\\gamma}=8.6\\times 10^3eV=\\frac{hc}{\\lambda}=\\frac{h}{\\lambda}c=p_{\\gamma}c\\\\\n\\implies p_{\\gamma}=\\frac{E_{\\gamma}}{c}=8.6keV\/c"Thus,
"p_{osmium}=-8.6keV\/c"Kinetic energy is
"K=\\frac{p^2}{2m}"Hence,
"K_{osmium}=2.088\\times 10^{-4}eV"
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