# Half-life

Roentgen’s discovery of an unknown ray called “x-ray” started a race for other scientist to know more about this ray. Marie Curie was one of them; she started with an ore called Pitchblende. Using her own way to purify this ore she was able to discover an element called Polonium in the honor of her homeland Poland. After checking the residue, she found out that the radiation emitted was very intense. It was around 700-900 times the radiation of Polonium. That element in the residue was radium. Both elements are radioactive. They are very unstable. All radioactive elements have half-life. This is the time it takes for a radioactive substance to decay half of its original mass.

Formula for half-life

$A_f=A_i(\frac{1}{2})^{\frac{T}{t}}$

Where :

$A_f$ – final amount after the time T

$A_i$ – initial amount of substance

$T$     – the elapsed time

$t$      – Half-life of substance

Sample Problem 1:

A radioactive substance has a half-life of 3 days. Find the amount of substance left after 9 days if initially there is 100 g of it?

[toggle title=”Solution”]

Using the formula:

$A_f=(100 g)(\frac{1}{2})^{\frac{9}{3}}$ $A_f=(100 g)(\frac{1}{2})^3$ $A_f=(100 g)(\frac{1}{8})$

$A_f=12.5 g$ [/toggle]

Sample Problem 2:

John conducted an experiment to determine the half-life of an unknown substance. He plotted the equation of his findings he was able to come up with an equation $A_f(T)=16(\frac{1}{2})^{\frac{T}{25}}$ where T is in years. The substance mass is in grams.

Determine the following:

a) Half-life in years

b) Amount of substance left after 100 years

[toggle title=”Solution”]

a) 25 years

b) using the formula,

$A_f(T)=16(\frac{1}{2})^{\frac{T}{25}}$ $A_f(T)=16(\frac{1}{2})^{\frac{100}{25}}$ $A_f(T)=16(\frac{1}{2})^4$

$A_f(T)=1 g$ [/toggle]

Sample Problem 3:

Chin observed that after 16 days the substance weight ¼ of its original weight. What is the half-life of the substance is days?

[toggle title=”Solution”]

$A_f=A_i(\frac{1}{2})^{\frac{T}{t}}$ $\displaystyle\frac{A_f}{A_i}=\displaystyle\frac{1}{2}^{\frac{T}{t}}$ $\displaystyle\frac{1}{4}=\displaystyle\frac{1}{2}^{\frac{21}{t}}$ $(\displaystyle\frac{1}{2})^2=\displaystyle\frac{1}{2}^{\frac{21}{t}}$ $2=\displaystyle\frac{21}{t}$ $t=\displaystyle\frac{21}{2}$

$t=10.5 days$ [/toggle]

### Dan

Blogger and a Math enthusiast. Has no interest in Mathematics until MMC came. Aside from doing math, he also loves to travel and watch movies.

### 3 Responses

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