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Atoms and Nuclei question

2015 · Shift 2 · Q55
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Atoms and Nuclei question

2015 · Shift 2 · Q55

JEE AdvancedPhysicsAtoms and NucleiMultiple correct+4 / −2
A fission reaction is given by 92236U→54140Xe+3894Sr+x+y_{92}^{236}U \to _{54}^{140}Xe + _{38}^{94}Sr + x + y92236​U→54140​Xe+3894​Sr+x+y, where x and y are two particles. Considering 92236U_{92}^{236}U92236​U to be at rest, the kinetic energies of the products are denoted by KXe,KSr,Kx(2MeV){K_{Xe}},{K_{Sr}},{K_x}(2MeV)KXe​,KSr​,Kx​(2MeV)  and Ky(2MeV)\text { and } \mathrm{K}_{\mathrm{y}}(2 \mathrm{MeV}) and Ky​(2MeV), respectively. Let the binding energies per nucleon of 92236U_{92}^{236}U92236​U, 54140Xe_{54}^{140}Xe54140​Xe and 3894Sr_{38}^{94}Sr3894​Sr be 7.5 MeV, 8.5 MeV and 8.5 MeV, respectively. Considering different conservation laws, the correct options is/are
  1. A
    x = n, y = n, Ksr = 129 MeV, KXe = 86 MeV
  2. B
    x = p, y = e −-−, Ksr = 129 MeV, KXe = 86 MeV
  3. C
    x = p, y = n, Ksr = 129 MeV, KXe = 86 MeV
  4. D
    x = n, y = n, Ksr = 86 MeV, KXe = 129 MeV
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