Question

Phosphorus-32 has a half life of 14.3 days. how many days will it take for radioactive sample to decay to 1/8 it's size?

Answer 1

The time required can be easily calculated by using the formula 

Nt/No=[1/2]^t/(t1/2) 
where
 Nt=1/8 
No=1 
t1/2=14,3 
t=....? 
by putting the given values in the formula discussed above:
 
(1/8)/1=[1/2]^t/14,3 
(1/8)=[1/2]^t/14,3 
(1/2)^3=[1/2]^t/14,3 -----> by eliminating 1/2 from both sides, 
3=t/14.3
 t=14,3*3 
t= 42.9 days. 

Answer 2

Answer: 43.3 days

Explanation:

Half-life of Phosphorus-32 = 14.3 days

First we have to calculate the rate constant, we use the formula :

$k=\frac{0.693}{14.3\text{days}}$

$k=0.048\text{days}^{-1}$

Now we have to calculate the days for radioactive sample to decay to 1/8 it's size

Expression for rate law for first order kinetics is given by:

$t=\frac{2.303}{k}\log\frac{a}{a-x}$

where,

k = rate constant  = $0.048\text{days}^{-1}$

t = age of sample  = ?

a = let initial size  = 1

a - x = amount left after decay process  = $\frac{1}{8}$

Now put all the given values in above equation, we get

$t==\frac{2.303}{0.048}\log\frac{1}{\frac{1}{8}}$

$t=43.3days$

Thus it will take 43.3 days for radioactive sample to decay to 1/8 it's size.