A fission reaction is given by U 92236 → Xe 54140 + Sr 3894 + x + y , where x and y are two particles. Considering U 92236 to be at rest, the kinetic energies of the products are denoted by K _ Xe , K _ Sr , K _ x (

Correct answer: A. x = n, y = n, K _ Sr = 129MeV, K _ Xe = 86MeV array ll U → Xe + Sr + & x & + 2 array array ll y 2 array Q = 4 + K _ Xe + K _ Sr ...(i) - Q = E _ B = 236 × 7.5 - 140 × 8.5 - 94 × 8.5 ∴ Q = 219 ...(ii) ∴ K _ Xe + K _ Sr = 215MeV Since, both x & y have same KE ∴ both particles should have same mass & lighter body will have higher KE.

JEE Advanced 2015Paper 2

Exam: JEE Advanced
Year: 2015
Paper: Paper 2
Subject: Physics
Chapter: Electromagnetic Waves

JEE Advanced 2015 · Paper 2 · Physics · Electromagnetic Waves

A fission reaction is given by U 92236 → Xe 54140 + Sr 3894 + x + y...

A fission reaction is given by

U 92236 \to Xe 54140 + Sr 3894 + x + y

, where

x and

are two particles. Considering

U 92236

to be at rest, the kinetic energies of the products are denoted by

{K}_{Xe}, {K}_{Sr}, {K}_{x} ( 2MeV )

and

{K}_{y}

\left(2MeV\right)

, respectively. Let the binding energies per nucleon of

U 92236 , Xe 54140

and

Sr 3894

be 7.5 MeV, 8.5 MeV, and 8.5 MeV respectively. Considering different conservation laws, the correct option(s) is (are)

Options

A $x = n, y = n, {K}_{Sr} = 129MeV, {K}_{Xe} = 86MeV$
B $x = p, y = {e}^{-}, {K}_{Sr} = 129MeV, {K}_{Xe} = 86MeV$
C $x = p, y = n, {K}_{Sr} = 129MeV, {K}_{Xe} = 86MeV$
D $x = n, y = n, {K}_{Sr} = 86MeV, {K}_{Xe} = 129MeV$

Correct answer

A. $x = n, y = n, {K}_{Sr} = 129MeV, {K}_{Xe} = 86MeV$

Step-by-step solution

\begin{array}{ll}U \to Xe + Sr + & x & + \\ 2\end{array}

\begin{array}{ll}y \\ 2\end{array}

<br >

Q = 4 + {K}_{Xe} + {K}_{Sr}

...(i)<br >

- Q = {E}_{B} = 236 \times 7.5 - 140 \times 8.5 - 94 \times 8.5

<br >

∴ Q = 219

...(ii)<br >

∴ {K}_{Xe} + {K}_{Sr} = 215MeV

<br >Since, both

&

have same KE<br >

∴$ both particles should have same mass & lighter body will have higher KE.

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