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Question: A fission reaction is given by \({}_{92}^{236}U \to {}_{54}^{140}Xe + {}_{38}^{94}Sr + x + y\), wher...

A fission reaction is given by 92236U54140Xe+3894Sr+x+y{}_{92}^{236}U \to {}_{54}^{140}Xe + {}_{38}^{94}Sr + x + y, where x and y are two particles. Considering 92236U{}_{92}^{236}U to be at rest, the kinetic energies of the products are denoted byKXe1,KSr1,Kx(2MeV) and Ky(2MeV){K_{X{e^1}}},{K_{S{r^1}}},{K_x}{\text{(2MeV) and }}{K_y}(2MeV), respectively. Let the binding energies per nucleon of 92236U{}_{92}^{236}U, 54140Xe{}_{54}^{140}Xe and 3894Sr{}_{38}^{94}Sr be 7.5 MeV, 8.5 MeV and 8.5 MeV, respectively. Considering different conservation laws, the correct option(s) is (are).

A. x=n,y=n,KSr=129MeV,KXe=86MeVx = n,y = n,{K_{Sr}} = 129MeV,{K_{Xe}} = 86MeV
B. x=p,y=e,KSr=129MeV,KXe=86MeVx = p,y = {e^ - },{K_{Sr}} = 129MeV,{K_{Xe}} = 86MeV
C. x=p,y=n,KSr=129MeV,KXe=86MeVx = p,y = n,{K_{Sr}} = 129MeV,{K_{Xe}} = 86MeV
D. x=n,y=n,KSr=86MeV,KXe=129MeVx = n,y = n,{K_{Sr}} = 86MeV,{K_{Xe}} = 129MeV

Explanation

Solution

This question can be answered by using newton’s second law. According to Newton's second law for a particular system rate of change of momentum will be equal to the external force acting on the system. Here the decay process is considered as an internal action.

Formula used:
Fext=dpsystemdt{F_{ext}} = \dfrac{{d{p_{system}}}}{{dt}}
K.E=p22mK.E = \dfrac{{{p^2}}}{{2m}}
Q=Δmc2Q = \Delta m{c^2}

Complete step by step answer:
Law of conservation of mass states that the mass can never be created nor destroyed, it is just converted from one form to the other. During the mass defect the entire mass which is lost should be appeared in some or the other manner. Here the mass will be appeared in the form of binding energy which will be denoted as B.E and its formula is given as
B.E=Δmc2B.E = \Delta m{c^2}
Where
Δm\Delta m is the difference in masses called as mass defect
‘c’ is the velocity of light
92236U54140Xe+3894Sr+x+y{}_{92}^{236}U \to {}_{54}^{140}Xe + {}_{38}^{94}Sr + x + y
For the given reaction difference in the binding energies of the reactants and products will be
236×7.5+140×8.5+94×8.5=219MeV- 236 \times 7.5 + 140 \times 8.5 + 94 \times 8.5 = 219MeV
So this difference will appear as the kinetic energy of the products.
Kinetic energy of x and y is given as 2MeV. So total it will be 4 MeV.
\eqalign{ & K + 4 = 219 \cr & K = 215 = 129 + 86 \cr}
All the given options satisfy this energy conservation. Now coming to the conservation charge on the left side is 92 and charge on the right side is 54+38=9254 + 38 = 92 which means x and y should not have any charge. Hence only A and D satisfies that.
We have K.E=p22mK.E = \dfrac{{{p^2}}}{{2m}}
If mass is less then kinetic energy must be more. Since the initial momentum is zero, xenon and strontium must have equal and opposite momentum to satisfy the momentum conservation to get the zero final momentum.
Hence the atom which is less mass should have more kinetic energy. So strontium should have more kinetic energy than xenon.
Only option A satisfies that.
Hence answer would be option A.

Note: Here if binding energy per nucleon of products is more than we can tell that reaction is more feasible one. Because in nature everybody tries to reduce its energy to attain stability. The mass lost in the reaction as the mass defect is converted to increase the binding energy of the products. That is the reason why nucleus has less mass than the individual nucleons.