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Problem 27 Medium Difficulty

Identify the missing nuclides in the following decays:
a) $^{212}_{83}\mathrm{Bi} \rightarrow ?+^{4}_{2} \mathrm{He}$
b) $^{95}_{36}\mathrm{Kr} \rightarrow ?+\mathrm{e}^{-}+\overline{\nu}$
c) $? \rightarrow_{2}^{4} \mathrm{He}+_{\mathrm{58}}^{140} \mathrm{Ce}$

Answer

a. _{83}^{212} B i \rightarrow_{81}^{208} T l+_{2}^{4} H e
b. \underset{36}{96} K r \rightarrow_{37}^{95} R b+_{-1}^{0} e+D
c. _{60}^{144} N d \rightarrow_{2}^{4} H e+_{58}^{140} C e

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Top Physics 103 Educators
Elyse G.

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Video Transcript

the number 27 for each one of these. I just need to complete the equation. So here I start with this business. Um, it has amassed number 28 to 12. And here again, tell her I lose. I lose an alpha particle. This is Alfa Decay here, so I'm losing, you know? Well, breaking off four. It's a man. So this will be to await on top on the bottom. I'm losing two protons. So this is 81. So I just look up on the periodic table and see what has an atomic number of 81. And that is Valium. C O report. Be Oh, this is beta decay. So when you're beta decay, one of the new trans splits into a proton and electron So here's the electron. So I didn't I didn't really change mass Number of all mass number Still 95 because just that neutron turned into a poem. But I do have one more proton supposed to be 37. So I look in the periodic table, see what has a temperament with 37 that's rubidium. Or be for the last one. Here it is Alfa decay again. So my total number of new clan 1 40 in that for on that. So it must have started with 1 44 and home in 48 2 more. It must have started with 50 pro tones. So looking to hear a table and see that's the and do no, you didn't.

University of Virginia
Top Physics 103 Educators
Elyse G.

Cornell University

Marshall S.

University of Washington

Farnaz M.

Other Schools

Zachary M.

Hope College