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(d) On a per unit mass basis, the previous result becomes (according to Eq. 4232)
020
23 10
3
.
.
mJ
85kg
J / kg = 2.3mGy.
=×
−
(e) Using Eq. 4231, (2.3 mGy)(13) = 30 mSv.
69. (a) Adapting Eq. 4221, we find
N
0
32
3
18
25 10
602 10
239
63 10
=
××
−
..
/
gm
o
l
g/mol
ch
c
h
(b) From Eq. 4215 and Eq. 4218,
( )
(
)
()
1/2
12h ln 2/ 24,100y 8760h/y
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 Spring '08
 Any
 Physics, Energy

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