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Calculate the energy released in the fission reaction

U235+n→Cs141+Rb93+2n

Here are some atomic and particle masses.

U235235.04392uRb9392.92157uCs141140.91964un1.00866u

Short Answer

Expert verified

The released energy in fission reaction is 181. MeV.

Step by step solution

01

Given data

The mass of U235,mu=235.04392u

The mass of R93b,mRb=92.92157u

The mass of C147s,mCs=140.91963u

The mass of n,n=1.00866

02

Determine the energy released in the fission reaction.

The expression to calculate released energy is given as follows.

Q=-∆mc2Q=(mu+mn-mCs-mRb-2mn)c2 ...(i)

03

Calculate the released energy in the fission reaction.

Calculate the released energy.

Substitute 235.04392u for m2=92.92157u, for mRb,140.91963u for mCs,1.00866u forn and 931.5MeV/u forc2 into equation (i).

Q=(235.04392u+1.00866u-140.91963u-92.92157-2×1.00866u)×931.5MeV/uQ=(235.04392+1.00866-140.91963-92.92157-2.01732)×931.5MeVQ=0.19406×931.5MeVQ=181 MeV

Hence the released energy in fission reaction is 181..MeV

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Most popular questions from this chapter

The fission properties of the plutonium isotope Pu239are very similar to those of U235. The average energy released per fission is 180 MeV. How much energy, in MeV, is released if all the atoms in 1.0 kg of purePu239 undergo fission?

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(a) If the reactor lasted for 200,000 y, at what average power level did it operate?

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