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Calculate the energy released in the re...

Calculate the energy released in the reaction, `""_(17)^(35)Cl + ""_(0)^(1)n to ""_(16)^(35)S + ""_(1)^(1)H`
` [(Cl-35) = 34.9688` amu (S-35) = 34.9690 amu, (H-1) = 1.0078 amu, `""_(0)^(1)n =1 .0087` amu]

A

0.6520 MeV

B

0.5620 MeV

C

0.2560 MeV

D

0.6250 Mev

Text Solution

Verified by Experts

The correct Answer is:
A

Mass defect `Deltam = [m(Cl) + m(n)] - [m(S) + m(H)]`
` = [34.9688 + 1.0087] - [34.9690 + 1.0078] = 0.0007` amu
Energy released ` = 0.0007 xx 931.48 = 0.6520` Mev
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