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1) If the half-life of 14056Ba is 12.75 days, how many
atoms of 14056Ba will be left after
9 weeks if there were initially 2.4 mol of atoms?
[9 weeks][7 days ][1 half
life ] = 4.94 halflives
[1 week][12.75 days]
[2.4 mol][6.02 x 1023
atoms] = [1.44
x 1024 atoms] = 4.7 x 1022 atoms
[1
mol
][24.94] [ 24.94
]
dividing by 24.94 is like multiplying by 14.94/24.94
or (1/2)4.94
2) How long will it take 3.8 x 1024 atoms of 10047Ag
to disintegrate to 2.93 x 1023 atoms
if the half life of 10047Ag is
24.6 seconds?
[original ]
= [3.8 x 1024 atoms ] = 2#halflives
= 12.97
[remaining] [2.93 x 1023
atoms]
halflives = [log 12.97]
= 3.7 halflives
[log 2 ]
[3.7 halflives][24.6 sec]
= 90.9 sec
[1 halflife]
3) Radioactive copper, 6429Cu, is found in quantities
exceeding pollution standards in the
sediments of a reservoir in a routine check on Monday.
The standard allows up to
14 ppm/cubic meter of sediment. On Monday, 59ppm/cubic
meter were measured.
The half life of 6429Cu is 12.7 hours.
When will the pollution level return to 14ppm?
[original ]
= [59 ppm] = 2#halflives = 4.2
[remaining] [14 ppm]
halflives = [log 4.2]
= 2.07 halflives
[log 2 ]
[2.07 halflives][12.7 hours]
= 26.3 hours later, or Tuesday at about the same time
[1 halflife ]
4) An unidentified corpse was discovered on 21 April at 7:00 AM. The
pathologist
discovered that there were 1.24 x 1037 atoms of 3215P
remaining in the victim's bones
and placed time of death sometime on 15 March. The half
life of 3215P is 14.28 days.
How much 3215P was present in the bones
at the time of death?
15 March --> 21 April is 38 days
[38 days][1 half
life ] = 2.66 halflives
[14.28 days]
[1.24 x 1037 atoms][22.66]
= 7.84 x 1037 atoms
multiplying by 22.66
is like dividing by 12.66/22.66 or (1/2)2.66
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