Calculate the energy released in the fission process. 92U235 + 0n = 56Ba141 + 36Kr92 + 3 0n1. Given M(U235) = 235.0439amu. M(Ba141) = 140.9129,M(Kr92) = 91.8973amu M(0n1) =1.0087amu

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

_{92} U^{235} + 0_{N} 1 \to_{5} 6 B a^{141} + K r^{92} + 3 X + 200 M e V

X

_{92} U^{235} +_{0} n^{1} \to_{38} S r^{90} + . . .

_{92} U^{235}

^{235} U

^{238} U

M (^{235} U) = 235.0439 a m u

M (n) = 1.0087 a m u

M (^{238} U) = 238.0508 a m u

M (^{236} U) = 236.0456 a m u

M (^{239} U) = 239.0543 a m u

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