92^(U^(235) mucleus absorbs a slow neutr...

`92^(U^(235)` mucleus absorbs a slow neutron and undergoes fission into `54^(X^139)` and `38^(sr^94)` nuclie The other particies produced in this fission process are

`1beta` and `1alpha`

`2 beta` and 1 neutron

2 neutrons

3 neutrons

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If an atom of _92^235U , after absorbing a slow neutron, undergoes fission to form an atom of -54^138 Xe and an atom of _38^94Sr , the other particles produced are

""_(92)U^(235) nucleus absorbs a neutron and disintegrates into ""_(54)Xe^(139) .""_(38)Sr^(94) and X. What will be the product X ?

""_(92)U""^(235) nucleus absorbes a neutron and disintegrate into ""_(54)Xe""^(139),""_(38)Sr""^(94) , and x neutrons x is

If ""_(92)U""^(235) nucleus absorbes neutron and disintegrates ""_(54)X""^(139).""_(38)Sr""^(95) and X. Then what will be the product X

The ._(92)^(235)U absorbs a slow neutron (thermal neutron) & undergoes a fission represented by (i) The energy release E per fission. (ii) The energy release when 1g of ._(92)^(236)U undergoes complete fission. Given : ._(92)^(235)U = 235.1175 am u (atom), ._(56)^(141)Ba = 140.9577 am u (atom), ._(36)^(92)Kr = 91.9264 am u (atom), ._(0)^(1)n = 1.00898 am u .

In a particular fission reaction, a ._92U^(235) nucleus captures a slow neutron. The fission products are three neutrons, a ._57La^(142) nucleus and a fission product ._ZX^A . What is the value of Z?

Uranium U_92^235 on bombardment with slow neutrons produces

Consider the case of bombardment of U^(235) nucleus with a thermal neutron. The fission products are Mo^(95) & La^(139) and two neutrons. Calculate the energy released by one U^(235) nucleus. (Rest masses of the nuclides are U^(235)=235.0439 u, ._(0)^(1)n=1.0087u, Mo^(95)=94.9058 u, La^(139).9061u) .

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