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Problem 32 Medium Difficulty

$\underset{28}{66} \mathrm{Ni}(\text { mass }=65.9291 \mathrm{u})$ undergoes beta decay to $^{66}_{29} \mathrm{Cu}$ (mass 5 65.928 9 u). (a) Write the complete decay formula for this process. (b) Find the maximum kinetic energy of the emerging electrons.

Answer

a. _{28}^{66} N i \rightarrow_{29}^{66} C u+_{-1}^{0} e+\overline{\nu}
b. 0.186 M e V

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Top Physics 103 Educators
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Rutgers, The State University of New Jersey

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University of Michigan - Ann Arbor

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Video Transcript

the number 32. We're too right in the dick a formula and find the kinetic energy of the electron. Well, the electrons, So their talent knows its beta decay. And you can tell that because these have the same mass number. Oh, So where do we start with this? So it's gonna be 66. 28 and I is gonna decay into 66. 29 c U plus an electron. And you can tell, since this has one more, um, Proton Here, this must be negative arrival electron, not a positron. And then for the energy, I just find energy difference here. So it started with this much, and then it would only has this much. So you can tell that's a difference of my subtract. I get that the missing mass is tweet. 02 use. I'm just going to convert that to its energy equivalence. So I know that one. You is the same as 9 31.5 mega electron volts. So the energy that's missing that is the kinetic energy of this electron. This point one 86 mega electron volts

University of Virginia
Top Physics 103 Educators
Christina K.

Rutgers, The State University of New Jersey

Andy C.

University of Michigan - Ann Arbor

Farnaz M.

Other Schools

Zachary M.

Hope College