Question 104: Atoms & Nuclei

Question

A fission reaction is given by $$_{92}^{236}U 	o _{54}^{140}Xe + _{38}^{94}Sr + x + y$$, where x and y are two particles. Considering $$_{92}^{236}U$$ to be at rest, the kinetic energies of the products are denoted by $${K_{Xe}},{K_{Sr}},{K_x}(2MeV)$$ $$
	ext { and } \mathrm{K}_{\mathrm{y}}(2 \mathrm{MeV})
$$, respectively. Let the binding energies per nucleon of $$_{92}^{236}U$$, $$_{54}^{140}Xe$$ and $$_{38}^{94}Sr$$ be 7.5 MeV, 8.5 MeV and 8.5 MeV, respectively. Considering different conservation laws, the correct options is/are

Accepted Answer

$${}_{92}^{236}U 	o {}_{54}^{140}Xe + {}_{38}^{94}Sr + x + y$$
K
x
= 2 MeV, K
y
= 2 MeV, K
Xe
= ?, K
Sr
= ?
By conservation of charge number and mass number, x $$\equiv$$ y $$\equiv$$ n
B.E. per nucleon of $${}_{92}^{236}U = 7.5$$ MeV
B.E. per nucleon of $${}_{54}^{140}Xe$$ or $${}_{38}^{94}Sr = 8.5$$ MeV
Q value of reaction,
Q = Net kinetic energy gained in the process
$$ = {K_{Xe}} + {K_{Sr}} + 2 + 2 - 0 = {K_{Xe}} + {K_{Sr}} + 4$$ ..... (1)
As number of nucleons is conserved in a reaction, so Q = Difference of binding energies of the nuclei
$$ = 140 	imes 8.5 + 94 	imes 8.5 - 236 	imes 7.5 = 219$$ MeV ...... (2)
From eqns. (i) and (ii)
$${K_{Xe}} + {K_{Sr}} = 219 - 4 = 215$$ MeV ....... (3)
The linear momentum of a particle of mass m and kinetic energy K is given by $$p = \sqrt {2mK} $$. Since the masses and kinetic energies of x and y are very small in comparison to that of $${}_{54}^{140}Xe$$ and $${}_{38}^{94}Sr$$, we can neglect the linear momentum of these particles. Initially, the linear momentum of $${}_{92}^{236}U$$ is zero (at rest). Finally, the products $${}_{54}^{140}Xe$$ and $${}_{38}^{94}Sr$$ will move in opposite direction with equal linear momentum (by conservation of linear momentum). Thus,
$$\sqrt {2{M_{Xe}}{K_{Xe}}}  = \sqrt {2{M_{Sr}}{K_{Sr}}} $$, i.e.,
$$140{K_{Xe}} = 94{K_{Sr}}$$. ...... (4)
Solve equations (3) and (4) to get K
Xe
= 86 MeV and K
Sr
= 129 MeV.

Answer

x = n, y = n, Ksr = 129 MeV, KXe = 86 MeV

Answer

x = p, y = e$$-$$, Ksr = 129 MeV, KXe = 86 MeV

Answer

x = p, y = n, Ksr = 129 MeV, KXe = 86 MeV

Answer

x = n, y = n, Ksr = 86 MeV, KXe = 129 MeV

Atoms & Nuclei

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