Answer:
[tex]A(t)=A_{0}e^{\frac{ln(\frac{1}{2})}{22}t}[/tex]
Step-by-step explanation:
This half life exponential decay equation goes by the formula:
[tex]A(t)=A_{0}e^{kt}[/tex]
Where
[tex]k=\frac{ln(\frac{1}{2})}{Half-Life}[/tex]
Since half life is given as 22, we plug that into "Half-Life" in the formula for k and then plug in the formula for k into the exponential decay formula:
So,
[tex]k=\frac{ln(\frac{1}{2})}{Half-Life}\\k=\frac{ln(\frac{1}{2})}{22}[/tex]
Now
[tex]A(t)=A_{0}e^{\frac{ln(\frac{1}{2})}{22}t}[/tex]
third choice is correct.
