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JEE · NEET Previous Year Question

$In a hypothetical fission reaction_{92} U^{235} +_{0} n^{1} \to_{56} Y^{141} +_{36} Z^{92} + 3 R The identity of emitted particles (R) is:$

Topic: JEE PYQ

Answer: B. Full step-by-step solution below.

A $Electron$
B $Neutron$
C $γ -radiations$
D $Proton$

Correct answer: B

Solution

(b) Z in LHS = 92 Z in RHS = 56 + 36 = 92 A in LHS = 236 A in RHS = 141 + 92 = 233 So, 3 neutrons are released.

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