Half‐lifeexamples,CH101Fall2009 10/14/2009
Half‐lifeexamplecalculations
CH101Fall2009
BostonUniversity
Half‐lifecalculations
Fundamentalrelation
(1/2)
n
N
0
=N
1
Half life is n for N/N
0
=
1/2; independent of N
0
1
.
Half
life
is
n
for
N/N
0
1/2;
independent
of
N
0
.
2. GivenN/N
0
aftertimet,calculaten,andthen
t
half
=t/n
3. Givenn,calculateN/N
0
,thefractionremaining
aftert=nt
half
Half‐lifecalculations
(1/2)
n
N
0
=N
Half‐lifeis33minutes.Whatfractionremainsafter
75minutes?
Strategy:Usen=75/33andthen
fractionremaining=N/N
0
=(1/2)
n
Answer:n=75/33=2.3,(1/2)
n
=0.21(21%)
Half‐lifecalculations
(1/2)
n
N
0
=N
Half‐lifeis11days.Howlongtodecayby10%?
Strategy:UseN/N
0
=0.90,solveN/N
0
=(1/2)
n
for
n=time/t
half
,andsotime=nxt
half
Answer:log[(1/2)
n
]=−nlog(2)=log(N/N
0
)
son=−log(N/N
0
)/log(2)
=−log(0.90)/log(2)=0.152,
hence,time=0.152x11days=1.7days
Half‐lifecalculations
(1/2)
n
N
0
=N
Asampledecaysby25%in15minutes.WhatIsits
half‐life?
Strategy:UseN/N
0
=0.75andn=15min/t
half
,
solveN/N
0
=(1/2)
n
forn=time/t
half
,
andsot
half
=time/n
Answer:n=−log(N/N
0
)/log(2)
=−log(0.75)/log(2)=0.415,
hencet
half
=15/0.415minutes=36minutes
Half‐lifecalculations
Fundamentalrelation
(1/2)
n
N
0
=N
CPSlesson:Half‐lifeexamples
