When a slow moving neutron collides with Uranium 235 atom $(\underset{92}{235} U)$ , 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 of reaction is 37 x K MeV. The value of K is _________. Use following data.
$m (U^{235}) = 235.0439 u, m (n) = 1.0087 u,$
$m (Z r^{99}) = 98.9165 u, m (T e^{134}) = 133.9115 u .$

Question

When a slow moving neutron collides with Uranium 235 atom $(\underset{92}{235} U)$ , 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 of reaction is 37 x K MeV. The value of K is _________. Use following data.

$m (U^{235}) = 235.0439 u, m (n) = 1.0087 u,$

$m (Z r^{99}) = 98.9165 u, m (T e^{134}) = 133.9115 u .$

Infinity Learn · Accepted Answer

Δ$$M=mU235+mn1$$−$$mZr99+m(Te)134+3mn1=0.1985So$$, Q value of reaction $$isQ=$$ΔM×$$931.5=185MeV=37$$×$$5MeV$$

Binding energy - Nuclei

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$Δ M = \{m (U^{235}) + m (n^{1})\}$ $- \{m (Z r^{99}) + m (T e)^{134} + 3 m (n^{1})\} = 0.1985$
So, Q value of reaction is
$Q = Δ M \times 931.5 = 185 MeV = 37 \times 5 MeV$

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Binding energy - Nuclei

Remember concepts with our Masterclasses.

ΔM=mU235+mn1−mZr99+m(Te)134+3mn1=0.1985So, Q value of reaction isQ=ΔM×931.5=185MeV=37×5MeV

Ready to Test Your Skills?

free study material

$Δ M = \{m (U^{235}) + m (n^{1})\}$ $- \{m (Z r^{99}) + m (T e)^{134} + 3 m (n^{1})\} = 0.1985$
So, Q value of reaction is
$Q = Δ M \times 931.5 = 185 MeV = 37 \times 5 MeV$