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)
see full answer
Your Exam Success, Personally Taken Care Of
1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.
An Intiative by Sri Chaitanya
a
x=n,y=n,KSr=129 MeV, KXe=86 MeV
b
x=p,y=e−,KSr=129 MeV, KXe=86 MeV
c
x=p,y=n,KSr=129 MeV, KXe=86 MeV
d
x=n,y=n,KSr=86 MeV, KXe=129 MeV
answer is A.
(Unlock A.I Detailed Solution for FREE)
Best Courses for You
JEE
NEET
Foundation JEE
Foundation NEET
CBSE
Detailed Solution
From conservation laws of mass number and atomic number, we can say that x=n, y=n x= 01n,y= 01nFrom conservation of momentumPxe=PsrFrom K=P22m⇒K∝1mSince, msrmxe⇒KsrKxe∴Ksr=129MeV,Kxe=86MeV