Question

If a 12.0-g sample of radon-222 decays so that after 19 days only 0.375 g remains, what is the half-life of radon-222? 9.50 days 4.75 days 3.80 days 0.594 days

Answer 1

$A_{f}=A_{0}*(1/2)^{ \frac{t}{h}} A(f) - final amount, A(0) -initial amount,t - time, h- half-life. \\ \\0.375=12.0*(1/2)^{ \frac{19}{h}} \\ \\(0.375/12.0)=(1/2)^{ \frac{19}{h}} \\ \\log(0.375/12.0)=log(1/2)^{ \frac{19}{h}} \\ \\ \frac{19}{h}*log(1/2)= log(0.375/12.0) \\ \\ \frac{19}{h}= \frac{log(0.375/12.0)}{log(1/2)} \\ \\ h= \frac{19log(1/2)}{log(0.375/12.0)} =3.80 (days) \\ \\ h=3.80(days)$

Answer 2

Answer:

3.80 Days

Look at the above comment for the math.