When .92U^235 undergoes fission by absor...

When `._92U^235` undergoes fission by absorbing `._0n^1` and `._56Ba^144` and `._36Kr^89` are formed. The number of neutrons produced will be

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The number of neutrons released when ._92 U^235 undergoes fission by absorbing ._0 n^1 and (._56 Ba^144 + ._36 Kr^89) are formed, is.

Two other possible ways by which .^(235)U can undergo fission when bombarded with a neutron are (1) by the release of .^(140)Xe and .^(94)Sr as fission fragments and (2) by the release of .^(132)Sn and .^(101)Mo as fission fragments. In each case, neutrons are also released. Find the number of neutrons released in each of these events.

In equation ._(92)U^(235) + ._(0)n^1 to ._(56)Ba^(144) + ._(36)Kr^(89) + X : X is-

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) 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

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?

When U-235 undergoes fission, 0.1% of its mass is converted into energy. Then amount of energy released during the fission of 1kg of uranium-235 will be

The most important use of studying nuclear fission and fusion process was to harness nuclear energy for useful purpose of humanity. Nuclear reactor is the machine designed to harness nuclear energy to electricity. The basic principle involved is fission of uranium to krypton and barium by slow moving neutrons as in following ""_(0)n^(1)+""_(92)^(235)Uto""_(32)^(96)Kr+""_(56)^(141)Ba+3_(0)n^(1)+Q(200MeV) reaction: But the difficulty is that one neutron striking with ""_(92)^(235)U produces three (2.6 on an average) neutrons which can cause further fission The neutron intensity available for further fission is controlled by measuring reproduction factor defined as the average number of neutrons from each fission the causes further fission. Maximum value of K 2.6. For optimum operation of nuclear reactor, the value of Kis mentioned in Column-1 for various stages in Columu-Il match the correct columns : {:("Column-I","Column-II"),((A)K=1,(p)"Uncontrolled chain reaction"),((B)Klt1,(q)"Critical"),((C)Kge1,(r)"Sub critical"),((D)K~~25,(s)"Super critical"):}

If 200 MeV energy is released in the fission of a single U^235 nucleus, the number of fissions required per second to produce 1 kilowatt power shall be (Given 1 eV = 1.6 xx 10^-19 J ).

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