What type of radioactive decay will the isotopes 13B and 188Au most likely undergo? a. Beta emission, positron emission
b. Beta emission, beta emission
c. Positron emission, beta emission
d. Positron emission, positron emission

Radioactive decay refers to the loss of energy by an unstable atomic nucleus as radiation. The substance having unstable nuclei is radioactive.

The correct answer is:

Option B. Beta emission, beta emission

The radioactive decay of 13B and 188 Au can be explained as:

1. Neutron proton ratio refers to the particular nuclei that will undergo what type of radioactive decay.

2. In the B-13 element, the N/P ratio is beyond the stability of the atom, thus, it will emit beta emission to reduce the N/P ratio.

3. Similarly, the N/P ratio of the Au-188 is above the stability of the atom, thus, it undergoes the beta emission to reduce the N/P ratio.

Thus, the correct answer is Option B.

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b. Beta emission, beta emission

Explanation:

A factor to consider when deciding whether a particular nuclide will undergo this or that type of radioactive decay is to consider its neutron:proton ratio (N/P).

Now let us look at the N/P ratio of each atom;

For B-13, there are 8 neutrons and five protons N/P ratio = 8/5 = 1.6

For Au-188 there are 109 neutrons and 79 protons N/P ratio = 109/79=1.4

For B-13, the N/P ratio lies beyond the belt of stability hence it undergoes beta emission to decrease its N/P ratio.

For Au-188, its N/P ratio also lies above the belt of stability which is 1:1 hence it also undergoes beta emission in order to attain a lower N/P ratio.

the electron configuration that you gave me has 16 electrons ergo it has 16 protons ergo it is the element sulfur.

i believe the answer is true.