RADIOACTIVE DECAY AND HALF-LIVES

BY NICOLE, UMBERTO,PIETRO, MATTIA AND CHIARA

How can we understand the stability of an atom with

math?

An atom is formed by four components : nucleus which is surrounded by protons, neutrons and electrons which rotate around the nucleus.

The stability of the atoms depends from the number of protons, electrons or neutrons that they contain

If a nucleus has an atomic number which is 0≤Z≤20, a stable atom can be formed if the amount of protons is equal to the quantity of the neutrons( in math, you would use the symbol Z in order to express this type of data).

If a nucleus has a number which is Z>20, it can be stable only if the number of neutrons of the atom is superior than the number of the protons.

When an atom is not electrically charged, it is defined as a neutral atom. The neutral atoms are the ones in which Z=n or n° ( n or n° is used to indicate the number of neutrons).

If a nucleus in an atom has a number in which Z>84, it means that the atom is unstable in contrast to the other ones. So, it is considered radioactive.

How is the atomic number written on the table of elements?

What is half-decay?

As we know, radioactive substances are formed by atoms in which their

nucleus is charged more than 84. In terms of radioactivity, half-life is

the time that each atom in a radioactive substance takes in order

to be reduced by half.

The elements colored in blue are stable elements

The half-life can happen even in 4 million years and those elements just release little amount of radiation. So, they are not harmful as other elements

The half-decay can happen in 800 years and 34000 years. Those substances can be harmful for a person’s health

The half-decay can happen from a day to 103 years and it causes problems to health

The half-decay can happen in a day or in a couple of minutes and they create serious and dangerous problems to a person’s health

The half-decay of those elements is not defined as they are extremely harmful

HOW TO USE MATH IN ORDER TO DESCRIBE AN ATOM’S

ELEMENT?Important aspects to describe an atom which are written using mathematics

Z is the atomic number

U or Da is used to indicate the atomic mass

Melting point

Boiling point

Number of protons/electrons

Number of neutrons written as n°

Density

Principal quantum number written as n. You should write from how many to how many levels.

EXAMPLE OF AN ELEMENT DESCRIPTION

Element: Astanine (At)

Z =85

Da = 210 amu

Melting Point: 302.0 °C

Boiling Point: 337.0 °C

e− =85

p+ = 85

n°= 125

The density is unknown

n= 1≤n≤6

n=1 e− = 2

n=2 e− = 8

n=3 e− = 18

n=4 e− = 32

n=5 e− = 18

n=6 e− = 7

WHAT IS THE PRINCIPAL QUANTUM NUMBER?

Quantum numbers describe the different existing atomic orbitals. Orbitals are regions in an atom in which there is the probability to find the electron we want to investigate on and to know in which region it can be

The principal quantum number is the number in a quantum mechanism which define the distance between an electron and a nucleus. This number is written as n. It can be any positive and real number but not zero.

This is the representation of astatine which is a radioactive element because its atomic number is more than 84. So, Z=85 when we are talking about astatine. Here, we can see that the atom has 6 energy levels. So, a result for n can be

1≤n≤6.

How radioactive decay is connected to exponentials

•Ao • t(1/2)

•A= final amount; Ao = Initial amount (1/2) = this is the split factor after a half-life, for example, one kilo becomes 1/2 kilo after xtime

•We can can use this formula to plot a graph•The graph represents the longevity of an atomic substance

FEATURES OF THE GRAPH

Domain: x € R, x ≠ - ∞, x ≠ 0 Range: y € R, y ≠ - ∞VS: NOHS: NO

Intercepts: NO Asymptote: x = 0

How logarithms are connected to radioactive

decayM=M0e-kt

M = remaining mass —> 6kg

MO = initial mass —> 1Okg

e = 2.71

k = decay constant

t = the time (years) -> 3months/O.25

Find time it takes to decay to 2kg

M= MOe-kt ==> 6= 1Oe-kO.25 (Since the time has to be in years, 3 months = O.25 years)

6= 1Ox2.7-kO.25

log2.7(6/1O)= -k x O.25

lnO.6= -k x O.25 (ln is like writing loge)

lnO.6/-0.25= k

k= 2.O433O2495

2kg= 1Oe-2.O433O2495 x t

2/1O= e-2.O433O2495 x t ==> O.2=e-2.O433O2495 x t

ln(O.2)= -2.O433O2495 x t

ln(O.2)/-2.O433O2495= t

t= O.788 years —> 287.62 days

Try it Yourself

Find time it takes to decay to 5kg

5= 1Oe-2.O433O2495 x t

5/1O= e-2.O433O24O95 x t

ln(O.5)= -2.O433O2495 x t

ln(O.5)/-2.O433O2495= t

t= O.399 years —> O.4 years = 146 days

after 500 years, a number of radio-226 decayed to 80.4% of its original mass. Find the half-life of the radio-226.

First we must consider k. We are given that after 500 years, the present quantity is equal to 80.4% of its original mass.

That is, when t = 500, A = 0.804 A0. By replacing values in the formula for exponential decay,

we obtain:0.804 A0 = A0ek (500).

Dividing through from A0 gives us 0.804 = e500k

which is an exponential equation.

To solve this equation, we take natural logs (ln) of both sides. (Even the common registers could be used). ln (0.804) = ln (e500k)

We know that ln (e500k) = 500k from the deletion properties of ln and e (see Logarithms). So the equation becomes ln (0.804) = 500k

k = (ln 0.84) / 500.

This is the exact solution; the natural register with a calculator to obtain the decimal approximation k = -0.000436.

ln (1/2) /(-0.000436)

or t = 1590. The half-life is about 1590 years.

Since I now know k, you can write the formula (function) for the amount of radio present at time t as

A = A0 and -0.000436 t.

Now you can finally find the half-life. We set A = 1/2 A0 and solve for t.

(1/2) A0 = A0 and -0.000436 t

By dividing again for A0, we get: 1/2 = e-0.000436 t.

To solve for t, take the natural registers: ln (1/2) = ln [e-0.000436 t].

Then apply the delete property for the returns of the logarithms ln (1/2) = -0.000436 t

So t = ln (1/2) /(-0.000436)

or t = 1590. The half-life is about 1590 years.

• The scientist received 200g of radioactive material and 24 days later 11g of chemical material are left remained after using it in

an experiment. What is the half life of this chemical material?

• A company has an investment of $8000 grows at the rate 9.7% per year, compounded annually. How long will it take for the investment to be worth $9500?