   Chapter 32, Problem 45PE

Chapter
Section
Textbook Problem

(a) Calculate the energy released in the neutron−induced fission reaction n + 239 Pu → 96 Sr + 140 Ba+4 n ,given m ( 96 Sr ) = 95.921750   u and m ( 140 Ba ) = 139.910581   u .(b) Confirm that the total number of nucleons and total charge are conserved in this reaction.

To determine

(a)

The energy released in the neutron-induced fission reaction n+P239uS96r+B140a+4n given m ( S96r ) =95.921750 u and m( B140a )=139.910581 u?

Explanation

Given info:

m (S96r) =95.921750 um(B140a)=139.910581 u

Formula used:

Energy,

E=Δmc2

Calculation:

Change in mass,

Δm=mn+m(P239u)

To determine

(b)

Confirmation that the total number of nucleons and total charge are conserved in this reaction?

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