(a) beta emission from aluminum-30 Al He + Na Al → e + Si 13 Al + e Mg 13 Al - e + Si 30 (b) alpha emission from einsteinium-252 je + 100 Fm Es Es 252 99 He + ie + % Cf Es +e→ 252 248 Bk 252 Es 252 252 99 98 Cf (c) electron capture by molybdenum-93 He + 20 Zr je +% Nb Mo + e ie + 3 Te Mo - 92 Mo - 93 Nb 41 Mo

Chemistry
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Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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Chapter1: Chemical Foundations
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Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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(a) beta emission from aluminum-30
13 Al → He + Na
18 Al - je + Si
18 Al + je - 2 Mg
* Al e + Si
30
30
30
(b) alpha emission from einsteinium-252
252
Es
je + 100 Fm
252
4
Es
248
He +
Bk
97
Es e + Cf
Es + e →
252
252
99
252
98 Cf
(c) electron capture by molybdenum-93
Mo - He + 20 Zr
je + % Nb
Mo + e →
ie +3 Te
92
Mo -
93
Nb
41
Mo
(d) positron emission by phosphorus-28
O 28 P - He + Al
+ je Si
ie + Si
je+ 16 s
28
14
28
28
28
P.
Transcribed Image Text:(a) beta emission from aluminum-30 13 Al → He + Na 18 Al - je + Si 18 Al + je - 2 Mg * Al e + Si 30 30 30 (b) alpha emission from einsteinium-252 252 Es je + 100 Fm 252 4 Es 248 He + Bk 97 Es e + Cf Es + e → 252 252 99 252 98 Cf (c) electron capture by molybdenum-93 Mo - He + 20 Zr je + % Nb Mo + e → ie +3 Te 92 Mo - 93 Nb 41 Mo (d) positron emission by phosphorus-28 O 28 P - He + Al + je Si ie + Si je+ 16 s 28 14 28 28 28 P.
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