- T1/2 = Half Life
- T = Elapsed Time
- Amt B = Beginning Amount
- Amt E = Ending Amount
Read the problem carefully and list out the values given.
Step 2 :
Substitute all the given values in the corresponding formula to find out the unknown value.
$T_{\frac{1}{2}} = \frac{T \times log2}{log\frac{Amt\ B}{Amt\ E}}$
The concentration of the beginning amount is 0.229 and the concentraton of ending amount is 0.282. Calculate the t1/2 in which the elapsed time is 280 sec.
Step 1 :Given data
Beginning amount = 0.229
Ending amount = 0.282
T (elapsed time) = 280 sec
Step 2 :Substitute the values in the formula to find out the t1/2.
$T_{\frac{1}{2}} = \frac{T \times log2}{log\frac{Amt\ B}{Amt\ E}}$
$T_{\frac{1}{2}} = \frac{280 \times log2}{log\frac{0.282}{0.229}}$
T1/2 = 932.25 sec
Answer :The answer is T1/2 = 932.25 sec.
For a radioactive decay the half life is given as 438 sec and the concentration of the beginning amount is 0.864 and the concentration of the ending amount is 0.897. Calculate the time elapsed in the decay.
Step 1 :Given data
T1/2 = 438 sec
Beginning amount = 0.864
Ending amount = 0.897
Step 2 :Substitute the values in the corresponding formula.
$T_{\frac{1}{2}} = \frac{T \times log2}{log\frac{Amt\ B}{Amt\ E}}$
$438 = \frac{T \times log2}{log\frac{0.897}{0.864}}$
T = 23.686 sec
Answer :The answer is T (elapsed time) = 23.686 sec