Use the graph of binding energy to estimate the total energy released if a nucleus with mass number 240 fissions into two nuclei with mass number 120.

Short Answer

Expert verified

The amount of energy released in the nuclear reaction involves when a nucleus of mass number 240 decays into two nucleus of each has mass number 120 approximately is 168MeV.

Step by step solution

01

Step.1.

The plot binding energy per nucleon verses mass number is given as the figure 42.6 in the text book. In this plot, the corresponding binding energy per nucleon for mass number 240 approximately is 7.8MeV. The corresponding binding energy per nucleon for mass number 120 approximately is 8.5MeV.

The total binding energy of nucleus of mass number 240 is,

E1=240(7.8MeV)=1872MeV

The total binding energy of two nuclei of each of mass number 120 is,

E2=120(8.5MeV)+120(8.5MeV)=2040MeV

02

Step.2

The amount of energy released in the nuclear reaction involves when a nucleus of mass number 240 decays into two nucleus of each has mass number 120 is,

ΔE=E2E1=2040MeV1872MeV=168MeV

Therefore, the amount of energy released in the nuclear reaction involves when a nucleus of mass number 240 decays into two nucleus of each has mass number 120 approximately is168MeV.

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Most popular questions from this chapter

We’ve noted that fewer than 10%of the known nuclei are stable (i.e., not radioactive). Because nuclei are characterized by two independent numbers, Nand Z, it is useful to show the known nuclei on a plot of neutron number Nversus proton numberZ.

The barium isotope 131Ba has a half-life of 12 days. A 250 μmg

sample of 131Ba is prepared. What is the mass of 131Ba after

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What kind of decay, if any, can occur for the nuclei in

FIGURE Q42.7?

Alpha decay occurs when an alpha particle tunnels through the Coulomb barrier. FIGURE CP42.63 shows a simple one-dimensional model of the potential-energy well of an alpha particle in a nucleus with A ≈ 235. The 15 fm width of this one-dimensional potential-energy well is the diameter of the nucleus. Further, to keep the model simple, the Coulomb barrier has been modeled as a 20-fm-wide, 30-MeV-high rectangular potential-energy barrier. The goal of this problem is to calculate the half-life of an alpha particle in the energy level E = 5.0 MeV. a. What is the kinetic energy of the alpha particle while inside the nucleus? What is its kinetic energy after it escapes from the nucleus? b. Consider the alpha particle within the nucleus to be a point particle bouncing back and forth with the kinetic energy you found in part a. What is the particle’s collision rate, the number of times per second it collides with a wall of the potential? c. What is the tunneling probability Ptunnel ? d. Ptunnel is the probability that on any one collision with a wall the alpha particle tunnels through instead of reflecting. The probability of not tunneling is 1 - Ptunnel. Hence the probability that the alpha particle is still inside the nucleus after N collisions is 11 - Ptunnel 2N ≈ 1 - NPtunnel , where we’ve used the binomial approximation because Ptunnel V 1. The half-life is the time at which half the nuclei have not yet decayed. Use this to determine (in years) the half-life of the nucleus.

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