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Neutrons from a source (perhaps the one discussed in the preceding problem) bombard natural molybdenum, which is 24 percent $^{98}$ Mo. What is the energy output of the reaction$98 \mathrm{Mo}+n \rightarrow^{99} \mathrm{Mo}+\gamma ?$ The mass of 98 $\mathrm{Mo}$ is given in

5.924 $\mathrm{MeV}$

Physics 103

Chapter 32

Medical Applications of Nuclear Physics

Nuclear Physics

Cornell University

Rutgers, The State University of New Jersey

University of Winnipeg

McMaster University

Lectures

02:51

In physics, wave optics is…

10:02

Interference is a phenomen…

03:04

Neutrons from a source (pe…

02:25

(a) Calculate the disinteg…

03:19

43.77. Calculate the energ…

01:31

Calculate the energy relea…

03:02

Find the energy released i…

02:08

00:54

Osmium $\stackrel{191}{76}…

00:41

Find the energy that is re…

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The purpose of producing 9…

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Complete the following fis…

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Find the number of neutron…

02:18

The purpose of producing $…

03:11

(II) Consider the fission …

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02:40

43.54. Thorium $_{90}^{230…

00:52

Molybdenum-99 is formed by…

04:53

in this problem. We want to know the energy output of a reaction, which is Mulet Dunham 98 plus neutrons resulting in well lived on, um, 99 plus some amount of gamma rays as an energy release. This is the energy output that we're looking for in this question. We know the known values off all these masses, so we can substitute them. 98 Mullah Dunham is 97 point 905406 a. M. Use atomic mass units. The mass of the neutron is one point. Oh 0866 5 a.m. Use the massive mulligan on 99 is 98.9 07711 am you this mass plus the unknown mass of the gamma rays. We will write as just gamma gamma rays. Energy is what we're trying to find the mass off. So the mass off gamma. Just by doing simple algebra, we can find that it is 0.636 am use. This is because 97.905406 plus 1.8665 is equal to 98.914 071 Thus, we can take 98 0.907711 And subtract this value that we found here 98.914071 To find the unknown value off the energy output. Taking the mass of the gamma rays, we can convert for mass to energy by using the standard conversion. 1 a.m. U. Is equal to 931 m E v e's all our troubles here we can write 0.636 atomic mass units multiplied by 931. Emmy V's is 5.9 24 m. E V s. And this is the value off the energy output of this reaction. By finding the mass off the energy that's released, we can convert from master Energy using this standard conversion.

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