# How do you determine alpha decay?

Feb 18, 2017

First know what is an alpha particle

Alpha decay is the decomposition of a nucleus of an element into a new nucleus of a different element and an alpha particle. Alpha particle is composed of 2 neutrons and 2 protons. Generally an alpha particle is referred to a helium nucleus since it has 2 neutrons and 2 protons but no electrons.

Charge of an $\alpha$ particle can be calculated

2protons = ${1}^{+} + {1}^{+} = {2}^{+}$
2neutrons =$0 + 0 = 0$

${2}^{+} + 0 = {2}^{+}$

Therefore an alpha particle has a charge of ${2}^{+}$ This photo indicates the alpha decay of uranium
The new nucleus that is formed which is Thorium is called the daughter nucleus and the nucleus which is decaying is the father nucleus which is uranium .

Now lets cone to the main question.

How to calculate the products of an alpha decay

First step In every alpha decay an alpha particle is formed though all alpha decay have different daughter nucleus . So first look at the father nucleus and list its number of protons and its atomic weight.

Its easy to understand when we solve an example . So lets take metal $' x '$ as the metal and its atomic weight as 14 and number of protons is 6
Step 2) Calculate the number of neutrons from the following equation
$\text{atomic weight - number of protons = number of neutrons}$
14 - 6protons = 8neutrons

Step 3) Now from number of neutrons subtract 2 and from number of protons subtract 2 as an alpha particle has 2 neutrons and 2 protons and in an alpha decay an alpha particle will always form in case of any any father nucleus.

For example (Please note that this is only an example)

protons = 6 so 6-2 = 4
neutrons= 8 so 8 -2 = 6

Step 3) After subtracting add the remaining protons and neutrons (4+6 = 10) 10 is the atomic weight of the new element nucleus.

So now this alpha decay can be represented as

""_6^14x = _4^10y+_2^4He

(You can also solve the equation by directly subtracting 4 from the atomic weight and 2 from number of protons but in chemistry we cannot give brief answers !!! )

If it were a real example you can check the element which has an atomic mass of 10. But as this is an example there is no element with 10 as the atomic weight

y metal is the new element nucleus formed.

Now lets solve a real alpha decay equation

Represent the alpha decay of Uranium

Atomic mass of seaborgium = 263
Protons = 106
neutrons = 263 - 106 = 157

106 - 2 = 104 protons
157- 2 = 155 neutrons

104 + 155 = 259

The element which has 259 as the atomic weight is rutherfordium.

So the equation is
""_106^263Sg --> ""_104^259Rf + ""_2^4He

""_2^4He can be also represented as $\alpha$ as this is the sign for alpha.

""_106^263Sg --> ""_104^259Rf + α