A fission reaction is given by _92^236U\to_54^140Xe\to_38^94Sr+x+y,where x and y are two particles.Considering _92^236U to be at rest, the kinectic energies of the produces are denoted by K_{Xe},K_2Me{Sr},K_{X}(2MeV)\text{and} K_{2MeV}. respectively, Let the binding energies per nucleon of _92^236U,_54^140Xe\text{and}_38^94Sr be 7.5 MeV,8.5 MeV and 8.5 MeV respectively. Considering different conservation laws, the correct option(s) is (are) [JEE Adv 2015 P2]

Correct Answer is : x=n,y=n,KSr=129MeV,KXe=86MeV

x=n,y=n,K_{Sr}=129MeV,K_{Xe}=86MeV

x=p,y=e^{-},K_{Sr}=129MeV,K_{Xe}=86MeV

x=p,y=n,K_{Sr}=129MeV,K_{Xe}=86MeV

x=n,y=n,K_{Sr}=86MeV,K_{Xe}=86MeV

JEE Advanced 2015 Paper 2

Section: Physics

Question : 16 of 60

Marks: +1, -0

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 kinectic energies of the produces are denoted by KXe,K2MeSr,KX(2MeV)and‌K2MeV. respectively, Let the binding energies per nucleon of 92236U,54140Xe‌and‌3894Sr be 7.5 MeV,8.5 MeV and 8.5 MeV respectively. Considering different conservation laws, the correct option(s) is (are)

[JEE Adv 2015 P2]

x=n,y=n,KSr=129MeV,KXe=86MeV
x=p,y=e−,KSr=129MeV,KXe=86MeV
x=p,y=n,KSr=129MeV,KXe=86MeV
x=n,y=n,KSr=86MeV,KXe=86MeV

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