The amount of time required by a substance to decrease by half, when the substance is undergoing decay, is called the

where,

$T_{\frac{1}{2}}$ = half life of a substance,

T = elapsed time,

AmtB = amount of substance required to decay at the beginning,

AmtE = amount of substance required to decay at the ending.

**half life**. Half life is basically used upon the radioactive substance which undergo exponential decay. The**half life calculator**will help to find the half life in exponential decay. The formula to calculate the half life is given as,$T_{\frac{1}{2}}$ = $\frac{T \times log2}{log\frac{AmtB}{AmtE}}$

where,

$T_{\frac{1}{2}}$ = half life of a substance,

T = elapsed time,

AmtB = amount of substance required to decay at the beginning,

AmtE = amount of substance required to decay at the ending.

**Step 1 :**From the problem, make a note of all the given values.

**Step 2 :**Substitute all the values into the half life formula and get the appropriate result accordingly.

Based on half life in exponential decay, some of the solved problems are given below.

### Solved Examples

**Question 1:**A radioactive substance is measured and after 120 days later, we find that it has 60% of the radioactivity it had when measured first. Determine the half life of the substance.

**Solution:**

Given : T = 120; AmtB = 1 ; AmtE = 0.60

Half life of the given substance is given by,

$T_{\frac{1}{2}}$ = $\frac{T \times log2}{log\frac{AmtB}{AmtE}}$

$T_{\frac{1}{2}}$ = $\frac{120 \times 0.30103}{log\frac{1}{0.60}}$

$T_{\frac{1}{2}}$ = $\frac{36.1236}{log(1.6667)}$

$T_{\frac{1}{2}}$ = $\frac{36.1236}{0.222}$

$T_{\frac{1}{2}}$ = 162.8

Therefore, the half life of the radioactive substance is 162.8 days.

**Question 2:**A chemistry teacher tells you to measure 100 grams of phosphorus-32. How much time will it take until it has 10 grams remain and if the half life is 75 days.

**Solution:**

Given : $T_{\frac{1}{2}}$ = 75; AmtB = 100 ; AmtE = 10 ; T = ?

Half life of the given substance is given by,

T = $T_{\frac{1}{2}}$ $\times$ $\frac{log\frac{AmtB}{AmtE}}{log2}$

T = 75 $\times$ $\frac{log\frac{100}{10}}{0.30103}$

T = 75 $\times$ $\frac{1}{0.30103}$

T = 249.144 days

Therefore, elapsed time of phosphorus-32 is 249.144 days.