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

(A) $\mathrm{x}=\mathrm{n}$, $\mathrm{y}=\mathrm{n}$, $K_{\mathrm{Sr}}=129\text{ MeV}$, $K_{\mathrm{Xe}}=86\text{ MeV}$

(B) $\mathrm{x}=\mathrm{p}$, $\mathrm{y}=\mathrm{e}^-$, $K_{\mathrm{Sr}}=129\text{ MeV}$, $K_{\mathrm{Xe}}=86\text{ MeV}$

(C) $\mathrm{x}=\mathrm{p}$, $\mathrm{y}=\mathrm{n}$, $K_{\mathrm{Sr}}=129\text{ MeV}$, $K_{\mathrm{Xe}}=86\text{ MeV}$

(D) $\mathrm{x}=\mathrm{n}$, $\mathrm{y}=\mathrm{n}$, $K_{\mathrm{Sr}}=86\text{ MeV}$, $K_{\mathrm{Xe}}=129\text{ MeV}$

JEE Advanced · 2015 · 24 May 2015 · Paper 2 · Modern Physics

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A fission reaction is given by

_{92}^{236} U \to_{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} (2 MeV)

and

K_{y} (2 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 option(s) is(are)

(A) $x = n$ , $y = n$ , $K_{Sr} = 129 MeV$ , $K_{Xe} = 86 MeV$ ✓ Correct
(B) $x = p$ , $y = e^{-}$ , $K_{Sr} = 129 MeV$ , $K_{Xe} = 86 MeV$
(C) $x = p$ , $y = n$ , $K_{Sr} = 129 MeV$ , $K_{Xe} = 86 MeV$
(D) $x = n$ , $y = n$ , $K_{Sr} = 86 MeV$ , $K_{Xe} = 129 MeV$

Answer: (A)

Solution

By conservation of mass number:

236 = 140 + 94 + A_{x} + A_{y}

, giving

A_{x} + A_{y} = 2

.
By conservation of atomic number:

92 = 54 + 38 + Z_{x} + Z_{y}

, giving

Z_{x} + Z_{y} = 0

.

Therefore, both x and y must be neutrons (n).

Q-value (energy released):

Q = 236 \times 7.5 - (140 \times 8.5 + 94 \times 8.5) = 1770 - 1989 = 219

MeV

This energy appears as kinetic energy:

Q = K_{Xe} + K_{Sr} + K_{x} + K_{y} = K_{Xe} + K_{Sr} + 2 + 2

K_{Xe} + K_{Sr} = 215

MeV.

By conservation of momentum (U at rest):

p_{Xe} = p_{Sr} + p_{x} + p_{y}

Since

K = \frac{p ^{2}}{2 m}

, and

m_{Sr} < m_{Xe}

, for momentum balance Sr must have higher kinetic energy.

Option (A) gives

129 + 86 = 215

MeV with Sr having higher KE, consistent with momentum conservation. ✓

Option (D) violates momentum conservation as Xe would have higher momentum than the sum of others.

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JEE Advanced · 2015 · 24 May 2015 · Paper 2 · Modern Physics

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