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Sagot :
Answer:
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.
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.
- For B-13, there are 8 neutrons and 5 protons, thus, the ratio is 1.6.
- For Au -188, there are 109 neutrons and 79 protons, thus, the ratio is 1.4.
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.
To know more about radioactive decay, refer to the following link:
https://brainly.com/question/14077220
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