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2 votes
How long will it take for a radioactive isotope with a decay constant of 0.15 (which means a half life of 4.6 days ) to decay to 5% of its original value?

2 Answers

6 votes
for half lives

A=P((1)/(2))^(t)/(h)

A=final amount
P=present amount
t=time in some units (this case it is days)
h=half life in days in th is case


so
given
we want 5% left
so A=5% of P or 5/100 of P or 1/20 of P or 0.05P

h=4.6

so

0.05P=P((1)/(2))^(t)/(4.6)
divide bth sides by P

0.05=((1)/(2))^(t)/(4.6)
take ln of both sides

ln(0.05)=ln((1)/(2))^(t)/(4.6)

ln(0.05)=((t)/(4.6))ln((1)/(2))

ln(0.05)=((t)/(4.6))ln((1)/(2))
divide both sides by
ln((1)/(2))

(ln(0.05))/(ln((1)/(2)))=(t)/(4.6)
times both sides by 4.6

(4.6ln(0.05))/(ln((1)/(2)))=t
use your calculator
19.889t
so after about 20 days
User Izk
by
9.1k points
1 vote
When we are dealing with an equation like this, we know that N=No exp(-kt), where k is the decay constant, 5% of its original value means N=(5/100)No
so, for t=t', N=(5/100)No, (5/100)No=Noexp(-kt'), (5/100)=exp(-kt')
Ln(5/100)=Lnexp(-kt')= - kt', -2.99 = -0.15t', so t' = 19.93
so the answer is E 19.97 days!
User Mike Scotty
by
8.5k points

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