Question

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

Answer: 11.3 days

Explanation: Radioactive decay follows first order kinetics.

Half-life of radon-222 = 3.823 days

$\lambda =\frac{0.693}{t_{\frac{1}{2}}}=\frac{0.693}{3.823}=0.18days^{-1}$

$N=N_o\times e^{-\lambda t}$

N = amount left after time t= 0.100 g

$N_0$ = initial amount= 0.750 g

$\lambda$ = rate constant=$0.18 days^{-1}$

t= time=?

$0.100=0.750\times e^{-0.18 days^{-1}\times tdays}$

$t=11.3days$