Question

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?

Answer 1

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!

Answer 2

For half lives
$A=P(\frac{1}{2})^\frac{t}{h}$

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
$0.05P=P(\frac{1}{2})^\frac{t}{4.6}$
divide bth sides by P
$0.05=(\frac{1}{2})^\frac{t}{4.6}$
take ln of both sides
$ln(0.05)=ln(\frac{1}{2})^\frac{t}{4.6}$
$ln(0.05)=(\frac{t}{4.6})ln(\frac{1}{2})$
$ln(0.05)=(\frac{t}{4.6})ln(\frac{1}{2})$
divide both sides by $ln(\frac{1}{2})$
$\frac{ln(0.05)}{ln(\frac{1}{2})}=\frac{t}{4.6}$
times both sides by 4.6
$\frac{4.6ln(0.05)}{ln(\frac{1}{2})}=t$
use your calculator
19.889t
so after about 20 days