Problem

(a) Background radiation due to ${ }^{226} \mathr…

04:19

Problem 22 Hard Difficulty

Naturally occurring 40 $\mathrm{K}$ is listed as responsible for 16 mrem/y of background radiation. Calculate the mass of 40 that must be inside the $55-\mathrm{kg}$ body of a woman to produce this dose. Each 40 $\mathrm{K}$ decay emits a 1.32 -MeV $\beta$ , and 50$\%$ of the energy is absorbed inside the body.

Answer

0.010 $\mathrm{g}$

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Physics 103

College Physics for AP® Courses

Chapter 32

Medical Applications of Nuclear Physics

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Video Transcript

no problem a 22 in chapter. So take two. And this problem, we want to know how much. But has there more? 40 is inside a woman to produce the amount off. 16 million rampart here. Oh, background that radiation. So 16 nearly 60 million ramp a year that you coz 0.16 nearly see road per year. So in one year among the mucks about with those eco's point 60 minutes rules per kilogram In here we have a used to that secret because draws per kilogram this is a definition off the dose. So for humans is 54 55 kilogram among a she absorbed it would be zero pone 16 times 55 and this equals 8.8 minute jewel's. So the number over the case in Y eah, it would be 8.8 million Jos 01 point That's 32 mega electoral rolls. This is the beta decay energy. Oh, potassium 40. Andi. We needed to come worth myriad walls to Omega Electoral votes or Meg electoral votes to millet rolls. We want to convert Munich Meg electrodes to meet. It does because it is easier to remember comes 10 to the negative 10 unit Jewel's and on the outside yes, and this gives a result of 4.17 times tend to the tense. So negativity we know that it equals the number of the case over time. So in one year is like we need to convert you two seconds and this result gives 13 to 2 pornos. Three p. Q. So we we already know that's the unit will be killed. Never need thio. It raises So we have more room for calculation. So the activity is defined as the total mess Or was the Moola? Meth comes Our Guardo's constant times look to always half lifetime This half lifetime is here is many, many long so long So from this we have that mass echoes 13 to 2 point the three This is pick you times forties a mola mazo potassium times the half life time you need to convert it to a seconds. Oh ah, A gradual number. We see 606.22 times 10 to the 23 comes along to. Besides, we also know that there is an information that only half is Ah Sabuda by the person So we need to time this by two. This is a total mess that existed in this moment. This gives a result. 0.0, one grass. This is a relatively small number.

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