Sales Toll Free No: 1-855-666-7446

Half Life Calculator

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

Steps

Back to Top
Steps to calculate the half life of the substance :

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.

Problems

Back to Top
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.