JEE 201 quesn doubt in ans

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The number of neutrons emitted when
²³⁵]U₉₂ undergoes controlled nuclear fission to
¹⁴²Xe₅₄ and
⁹⁰Sr₃₈ is ??
why is the ans given by JEE 4??how can it be 4??

#Chemistry #Physical Chemistry

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6 Answers

ratnadeep pramanik ·2011-03-08 06:26:46

Its 3 man......
its given rong i thnk.!!!!!

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Subhomoy Bakshi ·2011-03-08 12:29:56

235=142+90+3

bt the fission is initiated by 1 neutron...

thus, 235+1=142+90+4

hence the answer!!!

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AKHIL ·2011-03-09 02:56:55

the quesn says emitted , and not involved!!!!
:|

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Subhomoy Bakshi ·2011-03-09 03:52:27

it is emitted only!!

4 are emitted in the process!

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Ricky ·2011-03-09 04:05:31

₉₂ U ²³⁵ + ₀ n ¹ â†’ ₉₂ U ²³⁶ â†’ ₅₄ Xe ¹⁴² + ₃₈ Sr ⁹⁰ + 4 ₀ n ¹

Perhaps this'll clear your doubts .

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AKHIL ·2011-03-09 08:09:15

hmmm
ok
thnx..

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Basic Integral Calculus Operators Symbols Matrix Cases

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° ∂ ∞ α β γ δ θ λ μ ν ξ π ρ σ φ ω ε ƒ κ τ υ Γ Δ Λ Σ Π Ω

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JEE 201 quesn doubt in ans

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