`._(92)U^(238)` absorbs a neutron. The product emits an electron. This product further emits an electron. The result is

._(92)U^(238) on absorbing a neutron goes over to ._(92)U^(239) . This nucleus emits an electron to go over to neptunium which on further emitting an electron goes over to plutonium. How would you represent the resulting plutonium ?

A nucleus, absorbing a neutron, emits an electron to go over to neptunium which on futher emitting an electron goes over to plutonium. How would you represent the resulting plutonium?

U^238 emits

Consider the fission of ._(92)^(238)U by fast neutrons. In one fission event, no neutrons are emitted and the final end products, after the beta decay of the primay fragments, are ._(58)^(140)Ce and ._(44)^(99)Ru . Calculate Q for this fission process. The relevant atomic and particle masses are m(._(92)^(238)U)=238.05079 u m(._(58)^(140)Ce)=139.90543 u m(._(44)^(99)Ru)=98.90594 u

Consider the fission ._(92)U^(238) by fast neutrons. In one fission event, no neutrons are emitted and the final stable and products, after the beta decay of the primary fragments are ._(58)Ce^(140) and ._(44)Ru^(99) . Calculate Q for this fission process, The relevant atomic and particle masses are: m(._(92)U^(238))=238.05079u, m(._(58)Ce^(140))=139.90543 u, m(._(34)Ru^(99))=98.90594 u

Consider the fission of ""_(92)^(238)U by fast neutrons. In one fission event, no neutrons are emitted and the final end products, after the beta decay of the primary fragments, are ""_(58)^(140)Ce and ""_(44)^(99)Ru . Calculate Q for this fission process. The relevant atomic and particle masses are m(""_(92)^(238)U) =238.05079 u m( ""_(58)^(140)Ce ) =139.90543 u m(""_(44)^(99)Ru ) = 98.90594 u