Question

How long would it take for a sample of 222Rn that weighs 0.750 g to decay to 0.100 g? Assume a half-life for 222Rn of 3.823 days.

Answer 1

The initial Concentration is (N0) =0.705g
Final Concentraion is (N)= 0.100g
T(1/2)=3.823 days
      k(rate constant)=0.6932/T(1/2)
                               =0.1813 /days
     t=(2.303/k)xlog(N0/N)
      =(2.303/0.1813)xlog(0.750/0.100)
      = 12.702x0.87
      =11.051 days----->Answer

Answer 2

Formula we will be using:
(i) $Q_n=Q_0(0.5)^{ \frac{t}{n} }$,
where
$Q_n=$Final quantity
$Q_0=$Initial quantity
t=time ( same unit as that of half-life)
n=half-life
-------------------------------------------------------------

Given,
Final quantity =$Q_n$=0.100g
Initial quantity=$Q_0$=0.750g
Time= t(unknown)
Half-life=n=3.823 days


Plug in the above values in the formula:
$Q_n=Q_0(0.5)^{ \frac{t}{n} }$
$0.100=0.750(0.5)^{ \frac{t}{3.823} }$
Divide both sides by 0.750:
$\frac{10}{75}=0.5^ \frac{t}{3.823}$
Taking natural long on both sides:
$ln( \frac{10}{75})=ln(0.5)^ \frac{t}{3.823}$
Applying $ln(a)^x=xln(a):$
$ln( \frac{10}{75})= \frac{t}{3.823} ln(0.5)$
Dividing both sides by ln(0.5):
$\frac{ln( \frac{10}{75})}{ln(0.5)} = \frac{t}{3.823}$
Multiplying both sides by 3.823:
$3.823* \frac{ln( \frac{10}{75})}{ln(0.5)} =t$
11.113=t
t=11.113

It would take 11.113 days

Answer : 11.113 days