Chemistry 178: HW Set #7 Nuclear Chemistry Name: Materials in Chapters 21 and 14
ID: 590637 • Letter: C
Question
Chemistry 178: HW Set #7 Nuclear Chemistry Name: Materials in Chapters 21 and 14 may be useful. To receive full credit for this HW Set #7, you must: Due: Thursday, December 7, 2017 (12 pts maximum) TA/Section #: (1) Print out this page (2) Write your answers in the spaces provided on the right. Work may be shown in open areas. To receive full credit on Problem (4), you must show work. (3) Turn in this page only at the start of lecture into folders designated for your section. (1) (3 pts) Write the particle that is emitted in the following nuclear processes: (a) Gold-198 decays to mercury-198. (b) Radon-222 decays to polonium-218 (c) Potassium-40 decays to argon-40. Ans. Ans. (2) (3 pts) Write the daughter nucleus product, expressed as 2Sy, in the following nuclear processes: (a) Beta emission of manganese-56 (b Gallium-67 decays by electron capture (c) Potassium-38 decays by positron emission. Ans. Ans, (3) (3 pts) For each of the following pairs of isotopes, which one would you expect to be radioactive? Ans, Ans. Ans. (b) Fe or Co. (c) 20Hg or20Pb. (4) (3 pts) A sample of bone is subjected to radiocarbon dating. The current decay rate observed for its carbon-14 content is 11.2 disintegrations per minute per gram of carbon. What is the age (in years) of the bone? The half life for carbon-14 is 5.73x10 yrs., and the decay rate for carbon-14 in living organisms is 15.3 disintegrations per minute per gram of carbon.Explanation / Answer
(1) (a) 0-1e, 19879Au ---------> 0-1e + 19880Hg (beta decay)
(b) 42He, 22286Rn ------> 21884Po + 42He (alpha decay)
(c) 01e , 4019K -------> 01e + 4018Ar (positron emission )
(2) (a) 5624Cr, 5625Mn -----> 0-1e + 5624Cr
(b) 6730Zn , 6731Ga + 0-1e --------> 6730Zn
(c) 3818Ar , 3819K -------> 01e + 3818Ar
(3) (a) 81Br is radioactive (b) 58Co (c) 208Hg
(4) Let X years be the age of bone,
(11.2 / 15.3) = 1/2 * (X / 5.73*103)
Apply log on both sides and solve for X
log (11.2/15.3) = log 0.5 * (X / 5.73*103 )
X = 2578.6886 years ~ 2.578*103 years is the age of that bone
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