The cesium isotope137Csis present in the fallout from aboveground detonations of nuclear bombs. Because it decays with a slow(30.2y)half-life into137Ba, releasing considerable energy in the process, it is of environmental concern. The atomic masses of the Csand Baare 136.9058 uand, respectively; calculate the total energy released in such a decay.

Short Answer

Expert verified

The total energy released in such decay is 1.21 MeV .

Step by step solution

01

The given data

  1. The half-life of137Cs,T1/2(137Cs)=30.2y,
  2. The cesium isotope 137Csis decayed into barium isotope137Ba.
  3. The atomic mass of cesium isotope137Cs.m137Cs=136.9071u ,
  4. The atomic mass of the barium isotope137Ba.m137Ba=136.9058u ,
02

Understanding the concept of decay  

The disintegration energy, also known as the Q-value, is the energy that is absorbed or released when a nuclear reaction takes place. Thus, we can consider the energy released as the disintegration energy in the decay process of the cesium to barium nucleus.

Formula:

The disintegration energy of a nuclear reaction,

Q=(mparentnucleus-mdaughternuclei)c2 (i)

03

Calculation of the total released energy

Let m137Csbe the mass of one atom of 55137Csand m137Babe the mass of one atom of 55137Ba. The energy released is the change in mass due to the decay and can be given using equation (i) considering the disintegration energy as follows:

Q=m137Cs-m137Bac2=136.9071u-136.9058u931.5MeV/u=0.0013u931.5MeV/u=1.21MeV

Hence, the total energy released is 1.21 MeV .

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Most popular questions from this chapter

A typical kinetic energy for a nucleon in a middle-mass nucleus may be taken as 5.00MeV. To what effective nuclear temperature does this correspond, based on the assumptions of the collective model of nuclear structure?

Suppose the alpha particle in a Rutherford scattering experiment is replaced with a proton of the same initial kinetic energy and also headed directly toward the nucleus of the gold atom. (a) Will the distance from the center of the nucleus at which the proton stops be greater than, less than, or the same as that of the alpha particle? (b) If, instead, we switch the target to a nucleus with a larger value of Z,is the stopping distance of the alpha particle greater than, less than or the same as with the gold target?

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The maximum energy of the emitted positrons is 0.960 MeV. (a) Show that the disintegration energy Qfor this process is given by

role="math" localid="1661759171201" Q=(mC-mB-2me)c2

WheremCandmBare the atomic masses ofC11andB11, respectively, andmeis the mass of a positron. (b) Given the mass valuesmC=11.011434u,mB=11.009305uandme=0.0005486u, calculate Qand compare it with the maximum energy of the emitted positron given above. (Hint:LetmC andmBbe the nuclear masses and then add in enough electrons to use the atomic masses.)

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