A slow neutron strikes a nucleus of `._(92)^(235)U` splitting it into lighter nuclei of `._(56)^(141)Ba` and `._(36)^(92)Kr` along with three neutrons. The energy released in this reaction is ( The masses of uranium, barium and krypton of this reaction are `235.043933,140.917700` and `91.895400 u` respectively. The mass of a neutron is `1.008665 u`

In the nuclear chain ractio: ._(92)^(235)U + ._(0)^(1)n rarr ._(56)^(141) Ba + ._(36)^(92)Kr + 3 ._(0)^(1)n + E The number fo neutrons and energy relaesed in nth step is:

A slow neutron strikes a nucleus of ._(92)U^(235) splitting it into lighter nuclei of barium and krypton and releasing three neutrons. Write the corresponding nuclear reaction. Also calculate the energy released on this reaction Given m(._(92)U^(235))=235.043933 "amu" m(._(0)n^(1))=1.008665 amu m(._(56)Ba^(141))=140.917700 "amu" m(._(36)Kr^(92))=91.895400 amu

If three neutrons are added to nuclei of _92^235 U the new particles have an atomic number of

Calculate the energy released in the following reactions, the masses given are: ''_92^236U rarr _56^140Ba+_36^94Kr+2n

A neutron strikes a ""_(92)U^(235) nucleus and as a result ""_(36)Kr^(93) and ""_(56)Ba^(140) are produced with

A neutron strikes a ""_(92)U^(235) nucleus and as a result ""_(36)Kr^(93) and ""_(56)Ba^(140) are produced with

A slow neutron (n) is captured by a ._(92)U^(235) nucleus forming a highly unstable nucleus ._(92)U^(236**) (where ** ddenotes that the nucleus is in an excited state). The fission of the nucleus occurs by

If a ._(92)U^(235) nucleus upon being struck by a neutron changes to ._(56)Ba^(145) , three neutrons and an unknown product. What is the unknown product?