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When a slow moving neutron collides with Uranium 235 atom $({}_{92}^{235}U)$ , following reaction occurs.
$_{0}^{1} n +_{92}^{235} U ⟶_{92}^{236} U^{*} ⟶_{40}^{99} Zr +_{52}^{134} Te + 3_{0}^{1} n$
If kinetic energy of neutrons is neglected (as it is very small), then the Q-value of reaction is 37k MeV. The value of k is _________. Use following data.
$m (U^{235}) = 235.0439 u, m (n) = 1.0087 u,$
$m (Z r^{99}) = 98.9165 u, m (T e^{134}) = 133.9115 u .$

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Detailed Solution

$Δ M = \{m (U^{235}) + m (n^{1})\}$ $- \{m (Z r^{99}) + m (T e)^{134} + 3 m (n^{1})\} = 0.1985$

So, Q-value of reaction is

$Q = Δ M \times 931.5 ≃ 185 MeV = 37 \times 5 MeV$

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