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Chapter 11: Q39E (page 519)

How much Kinetic energy released and what is the daughter nucleus in theαdecay of polonium-210 ?

Short Answer

Expert verified

Kinetic energy release in alpha decay of polonium−210 is5.41 MeV , and daughter nucleus is role="math" localid="1658408913570" Lead−206.

Step by step solution

01

Given data

In alpha decay of polonium−210.

84210Po→24α+82206Pb

Atomic mass of84210Po,mPo0=209.982848u.

Atomic mass of82206Pb,mPb=205.97444u.

Atomic mass of 24α,mα=4.002603u.

02

Definition of Kinetic Energy

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

Where,miis the mass of parents’ nucleus,mfis the mass of the daughter nucleus, andcis the speed of light in vacuum.

03

Determine the Kinetic Energy released

Kinetic energy release,Q=(mi−mf)c2 .

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

Q=(mPo−mPb−mα)931.5MeVuQ=(209.982848u−205.97444u−4.002603u)931.5McVuQ=5.41MeV

Kinetic energy release in alpha decay of polonium−210 is 5.41 MeV, and daughter nucleus is Lead−206.

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Most popular questions from this chapter

(a) How much energy can be extracted by deuterium fusion from a gallon of sea water? Assume that an average D-D fusion yield is about per atom.

(b) A modem super tanker can hold9×107 gallons. How many "water tankers" would be needed to supply the energy need of greater Los Angeles, consuming electricity at a rate of about 20 GW, for 1 year? Assume that only20% of the available energy actually becomes electrical energy.

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(a) Calculate the ratio of the second term to first term forA=20 . What does it say about the surface nucleons?

(b) Repeat part (a), but forA=220 .

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