92^(U^(235)) nucleus absorbs a neutron a...

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

3 neutrons

2 neutrons

`alpha` particle

`beta` particle

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On neutron bombardment fragmentation of U-235 occurs according to the equation. 92^(U^(235)) + ^(0n^(1)) rarr 42^(Mo^(98)) + 54^(Xe^(136)) + x_(-1)^(e^(0)) + y 0^(n^(1)) Calculate the values of x and y.

On neutron bombardment fragmentation of U-235 occurs according to the equation 92^(U^(235)) + 0^(n^(1)) rarr 42^(Mo^(95)) + 57^(La^(139)) + x_(-1)^(e^(0)) + y_(0)^(n^(1)) Calculate the values of x and y.

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

Calculate the number of neutrons in the remaining atom after emission of an alpha particle from 92^(X^(238)) atom. also report the mass number and atomic number of the product atom.

A fission reaction is given by ._(92)^(236)U rarr ._(54)^(140)Xe + ._(54)^(140)Xe + ._(38)^(94)Sr + x +y , where x and y are two particles. Considering ._(92)^(236)U to be at rest, the kinetic energies of the products are denoted by K_(xe),K_(Sr),K_(x)(2MeV) and K_(y)(2Me V) , respectively. Let the binding energies per nucleon of ._(92)^(236)U, ._(54)^(140)Xe and ._(38)^(94)Sr be 7.5 MeV, 8.5 Me V and 8.5 MeV , respectively. Considering different conservation laws, the correct options (s) is (are)

The mass of nucleus ._(z)X^(A) is less than the sum of the masses of (A-Z) number of neutrons and Z number of protons in the nucleus. The energy equivalent to the corresponding mass difference is known as the binding energy of the nucleus. A heavy nucleus of mass M can break into two light nuclei of mass m_(1) and m_(2) only if (m_(1)+m_(2)) lt M . Also two light nuclei of massws m_(3) and m_(4) can undergo complete fusion and form a heavy nucleus of mass M ''only if (m_(3)+m_(4)) gt M ''. The masses of some neutral atoms are given in the table below. |{:(._(1)^(1)H,1.007825u,._(1)^(2)H,2.014102u,),(._(1)^(3)H,3.016050u,._(2)^(4)H,4.002603u,),(._(3)^(6)Li,6.015123u,._(3)^(7)Li,7.016004u,),(._(30)^(70)Zn,69.925325u,._(34)^(82)Se,81.916709u,),(._(64)^(152)Gd,151.91980u,._(82)^(206)Pb,205.97445u,),(._(83)^(209)Bi,208.980388u,._(84)^(210)Po,209.982876u,):}| The correct statement is

Which of the following is/are correct for nuclear reactor? a)A typical fission is represented by ""_(92)""^(235)U=""_(0)""^(1)nto""_(54)""^(140)Ba+""_(36)""^(93)Kr+ Energy b)Heavy water ( D""_(2)O ) is used as moderator in preference to orduinary water ( H""_(2)O ) because hydrogen may capture neutrons, while D would not do that c)Cadmium rods increse the reactor power when they go in and decrease when they go outwards d)Slower neutrons are more effective in causing fission than faster neutrons in the case of ""_(235)U

The mass of a nucleus ._(Z)^(A)X is less that the sum of the masses of (A-Z) number of neutrons and Z number of protons in the nucleus.The energy equivalent to the corresponding mass difference is known as the binding energy of the nucleus. A heavy nucleus of mass M can break into two light nuclei of masses m_(1) and m_(2) only if (m_(1)+m_(2)) lt M . Also two light nuclei of masses m_(3) and m_(4) can undergo complete fusion and form a heavy nucleus of mass M'. only if (m_(3)+m_(4)) gt M' . The masses of some neutral atoms are given in the table below: |{:(._(1)^(1)H ,1.007825u , ._(1)^(2)H,2.014102u,._(1)^(3)H,3.016050u,._(2)^(4)He,4.002603u),(._(3)^(6)Li,6.015123u,._(3)^(7)Li,7.016004u,._(30)^(70)Zn,69.925325u, ._(34)^(82)Se,81.916709u),(._(64)^(152)Gd,151.91980u,._(82)^(206)Pb,205.974455u,._(83)^(209)Bi,208.980388u,._(84)^(210)Po,209.982876u):}| Taking kinetic energy ( in KeV ) of the alpha particle, when the nucleus ._(84)^(210)P_(0) at rest undergoes alpha decay, is:

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