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Chapter 11: Q40E (page 519)

How much Kinetic energy released and what is the daughter nucleus in the β+ decay of nitrogen-13?

Short Answer

Expert verified

Kinetic energy release in β+decay is1.708 MeV and daughter nucleus is Carbon-13.

Step by step solution

01

Given data

The β+decay of nitrogen- .

role="math" localid="1658409778492" 713N→613C++10β++v

Atomic mass of 713N,mN=13.005738u.

Atomic mass of 613C,mC=13.003355u.

Atomic mass of electron,me=5.486×10−4u .

02

Definition of Kinetic Energy 

The kinetic energy released is, Q=(mi−mf)c2.

Where,mi is the mass of parents’ nucleus, mfis the mass of the daughter nucleus, and cis the speed of light in vacuum.

03

Determine the Kinetic Energy released

Kinetic energy release in β+decay is Q=(mparent−mdaughter−me)c2.

The conversion factor may be used to express, c2→MeV/u .

Q=(mpurent−mdaughter−me)931.5MeVuQ=(mN−mC−2me)931.5MeVuQ=[13.005738u−13.003355u−5.486×10−4u]931.5MeVuQ=1.708MeV

Kinetic energy release in β+decay is 1.708 MeV and daughter nucleus is Carbon- 13.

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Most popular questions from this chapter

Exercise 19 notes that the energy needed to remove a neutron fron helium-4 is 20.6 MeV

(a) Show that the energy required to remove a proton is 19.8 MeV.

(b) Why do these values disagree with the BE/nucleon value shown for helium-4 in Figure 11.14?

MRI relies on only a tiny majority of the nuclear magnetic moments aligning with the external field. Consider the common target nucleus hydrogen. The difference between the aligned and anti aligned states of a dipole in a magnetic field is2μ:B Equation (8-7) can be used to findμz for the proton. Provided that the correct mass and gyromagnetic ratio(gp=5.6) are inserted. Using the Boltzmann distribution, show that for a1.0T field and a reasonable temperature, the number aligned exceeds the number anti aligned by less than 1100%.

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(c) Determine the binding energy per nucleon using the semiempirical binding energy formula and discuss the result.

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