Question

Write the beta decay equation for the following isotope: 77/33 As ?

Answer 1

`._33^77As` `->` 0/-1e + `._34^77Se`

Explanation:

First, we need to find the decay mode of `"^77As`.
You can find that it is `beta` decay (I found this information online). Beta decay can be written as 0/-1e `or` 0/-1`beta`. It doesn't matter which you choose.

In natural decay equations (which use beta, alpha, positron, etc. decay), there is always one reactant. In this case, the reactant is As.
Decay modes always go on the products side of the equation.

Therefore, we have:

`1_33^77As` -> 0/-1e + _#

(I had to put the one there for format purposes.)

All we need is the last product. According to the Law of Conservation, we need mass, charge, and energy to be the same on both sides of the equation.

If our mass number is 77 on the reactant side, the total mass number must be 77 on the product side. Therefore, the mass of our unknown product is 77 (0+77=77).

We need our atomic number to be the same as well (33). We have a -1, so we need an atomic number of 34 (-1+34=33).

Since our atomic number is 34, we need the correct element that has an atomic number of 34. Looking at the periodic table, that element is Selenium (Se).

`1_33^77As` -> 0/-1e + `color(red)(1_34^77Se)`

Answer 2

Refer to the explanation.

Explanation:

We have arsenic-`77` undergoing beta decay. I assume that this will undergo beta-minus decay, in which an atom releases an electron and an antineutrino, while converting a neutron into a proton as well.

Thus, arsenic will gain one proton, and will become selenium.

So, we get:

`""_33^77As->""_(-1)^(color(white)(>)0)e^(-)+""_34^77Se+barv`