A fission reaction is given by $$92236U$$→$$54140Xe+3894Sr+x+y$$, where x and y are two particles. Considering $$92236U$$ to be at rest, the kinetic energies of the products are denoted by KXe,KSr,$$Kx(2MeV)$$ and $$Ky2MeV$$ respectively. Let the binding energies per nucleon of $$92236U$$,$$54140Xe$$ and $$3894Sr$$ be $$7.5$$ MeV, $$8.5$$ MeV and $$8.5$$ MeV , respectively. Considering different conservation laws, the correct $$option(s)$$ is (are)

Question

A fission reaction is given by  $$92236U$$→$$54140Xe+3894Sr+x+y$$,  where  x and  y are two particles.  Considering  $$92236U$$ to be at rest, the kinetic energies of the products are denoted by  KXe,KSr,$$Kx(2MeV)$$ and $$Ky2MeV$$ respectively.  Let the binding energies per nucleon of   $$92236U$$,$$54140Xe$$ and   $$3894Sr$$ be  $$7.5$$ MeV, $$8.5$$ MeV and $$8.5$$ MeV , respectively.  Considering different conservation laws, the correct $$option(s)$$ is (are)

Infinity Learn · Accepted Answer

From conservation laws of mass number and atomic number, we can say that $$x=n$$, $$y=n$$ $$x=$$ $$01n$$,$$y=$$ $$01nFrom$$ conservation of $$momentumPxe=Psr$$​From $$K=P22m$$⇒K∝$$1m$$​Since, msrmxe​⇒KsrKxe​∴$$Ksr=129MeV$$,$$Kxe=86MeV$$

Start JEE / NEET / Foundation preparation at rupees 99/day !!

Detailed Solution

Get Expert Academic Guidance – Connect with a Counselor Today!