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Questions

Calculate the ground state Q value of the induced fission reaction in the equation
$n +_{92} U^{23} \to Z {r_{40}}^{99} + T {e_{52}}^{134} + 3 n$
$m (n) = 1.0087 u; M (U^{235}) = 235.0439 u$
$M (Z r) = 98.916 u; M (T e^{134}) = 133.9115 u$

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a

184.84 MeV

b

200 MeV

c

130 MeV

d

300 MeV

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detailed solution

Correct option is A

Q=Δm×931[(235.0439+1.0087) −(98.916+133.9115+3.0261)×931]Q=184.84MeV

Similar Questions

What is the total energy emitted for given fission reaction

$_{0}^{1} n +_{92}^{235} U ⟶_{92}^{236} U ⟶_{40}^{98} Z r +_{52}^{136} T e + 2_{0}^{1} n$

The daughter nuclei are unstable therefore they decay into stable end products ${}_{42}^{98}M o and {}_{54}^{136}X e$ by successive emission of β-particles. Let mass of $_{0}^{1} n = 1.0087$ amu, mass of $_{92}^{235} U = 236.0526$ amu, mass of $_{54}^{136} Xe = 135.9170$ amu, mass of ${}_{42}^{98}M o$ = 97.9054

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