Find the value of x in the following nuclear reaction- 92U236- 56Ba141 - 36Krx - 3 0n1 - Q

The correct option is C 92 _ 92 U ^ 236 → _ 56 Ba ^ 141 +_ 36 Kr ^ x +3_ 0 n ^ 1 +Q From conservation of mass 236=141+x+3× 1 x=236 141 3 x=9 ...

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Nuclear Fission

Find the valu...

Question

Find the value of $x$ in the following nuclear reaction.
$_{92} U^{236} \to_{56} {B a}^{141} +_{36} {K r}^{x} + 3_{0} n^{1} + Q$

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{_{92} U}^{235} + {_{0} n}^{1} ⟶ {_{56} B a}^{141} + {_{36} K r}^{92} + 3 x + Q (e n e r g y)

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{_{92} U}^{236} ⟶ X^{117} + Y^{117} + {_{0} n}^{1} + {_{0} n}^{1}

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8.5 M e V

U^{236}

7.6 M e V

U^{235} +_{0} n^{1} ⟶_{56} B a^{141} +_{36} K r^{92}

_{92}^{235} U +_{0}^{1} n \to_{56}^{139} B a +_{36}^{94} K r + 3_{0}^{1} n

_{92}^{236} X \to_{56}^{140} Y +_{36}^{94} Y + n e u t r o n s

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