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In equation .(92)U^(235) + .(0)n^1 to .(...

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

A

`._(0)n^(1)`

B

3H

C

`3._(0)n^(1)`

D

none of these

Text Solution

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The correct Answer is:
C
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The reaction ._(92)U^(235) + ._(0)n^(1) rarr ._(56)Ba^(140) + ._(36)Kr^(93) + 3 ._(0)n^(1) represents

A nuclear reaction is represented by the following equation : ._(92)^(235)U + ._(0)^(1)n rarr ._(56)^(139)Ba + ._(36)^(94)Kr + xc + E (a) Name the process represented by this equation and describe what takes place in this reaction. (b) Identify the particle c and the number x of such particles produced in the reaction. (c ) What does E represent ? (d) Name one installation where the above nuclear reaction is utilised. (e) What type of bomb is based on similar type of reactions ?

When Uranium is bombarded with neutrons , it undergoes fission . The fission reaction can be written as : ""_(92) U^(235) + ""_(0) n^(1) to ""_(56) Ba^(141) + ""_(36) Kr^(92) + 3x + Q (energy) where three particles named x are produced and energy Q is released . What is the name of the particle x ?

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:

U-235 is decayed by bombardment by neutron as according to the equation: ._(92)U^(235) + ._(0)n^(1) rarr ._(42)Mo^(98) + ._(54)Xe^(136) + x ._(-1)e^(0) + y ._(0)n^(1) Calculate the value of x and y and the energy released per uranium atom fragmented (neglect the mass of electron). Given masses (amu) U-235 = 235.044 , Xe = 135.907, Mo = 97.90, e = 5.5 xx 10^(-4), n = 1.0086 .

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