Q.Disintegration rate of a sample is $10^{10}$ per hour at 20 hrs from the start. It reduces to $6.3\times10^9$ per hour after 30 hours. Calculate its half life and the initial number of radioactive atoms in the sample.

[!TLDR] $\lambda=-\frac{1}{10\text{hr}}\ln(6.3/10)=0.0462$/hr, giving $T_{1/2}=0.693/0.0462\approx15$ hr; back-extrapolating to t=0 gives $N_0\approx5.45\times10^{ …

Between t=20 hr and t=30 hr (an interval of 10 hr), the activity falls from $10^{10}$/hr to $6.3\times10^9$/hr, a ratio of $6.3\times10^9/10^{10}=0.63$. Using $A(t)=A_0e^{-\lambda t}$ over this interval, $0.63=e^{-\lambda(10)}$, so $\lambda=-\frac{\ln(0.63)}{10}=\frac{0.4620}{10}=0.0462$ per hour. The half-life follows as $T_{1/2}=\frac{0.693}{0.0462}\approx15.0$ hours. For the initial (t=0) population, first back-extrapolate the activity to t=0 from the t=20 hr value: $A_0=A(20)\,e^{\lambda(20)}=10^{10}\times e^{0.0462\times20}=10^{10}\times e^{0.924}\approx2.519\times10^{10}$ per hour, which converts to $\approx6.997\times10^6$ per second. The i …

Use the ratio of the two given activities over the known 10-hour gap to find $\lambda$ directly, get $T_{1/2}$ from $\lambda$, then back …

Treating the given times (20 hr, 30 hr) as measured from a DIFFERENT reference than the sample's own t=0, instead of correctly using their 10-hour DIFFERENCE to find $\l …