Answer : The half-life of radium-230 is, 1 hr 33 min
Solution : Given,
Initial amount of radium-230 = 2.00 mg
Amount left after time, 't' = 0.25 mg
Time = 4 hr 39 min = [tex]4\times 60+39=279min[/tex] (1 hr = 60 min)
Rate law expression for first order kinetics :
[tex]N=N_o\times e^{-\lambda t}[/tex]
Taking 'ln' on both the sides, we get
[tex]\ln(\frac{N}{N_o})=-\lambda t[/tex]
where,
N = amount left after time t
[tex]N_0[/tex] = initial amount
[tex]\lambda[/tex] = rate constant
t = time
Now put all the given values in the above expression, we get
[tex]\ln(\frac{0.25}{2})=-\lambda \times (279min)[/tex]
By rearranging the terms, we get
[tex]\lambda=0.00745min^{-1}[/tex]
Radioactive decay follows first order kinetics.
[tex]t_{\frac{1}{2}}=\frac{0.693}{\lambda}[/tex]
[tex]t_{\frac{1}{2}}=\frac{0.693}{0.00745min^{-1}}=93.020min=1hr33min[/tex]
Therefore, the half-life of radium-230 is, 1 hr 33 min
