Answer:
B. There is a very large percentage of C-12.
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us to realize that, since the average atomic mass is 12.01 amu, then the C-12, with an atomic mass of 12.000 am prevails over C-13 with an atomic mass of 13.003 amu as long as the average is nearer to the former.
In such a way, the answer will be B. There is a very large percentage of C-12.
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Because it has atomic no. = 1 (according to modern periidic table)
<u>¹⁴₇N</u><u> </u>is the more stable isotope
<h3>
Briefly explained</h3>
We have ¹⁴₇N which has a neutron to proton ratio of one, and we look at ¹⁸₇N which has a neutron to proton ratio of 1.57 Again, you look at table 24 to and you see the atomic number of seven and there is really no stable isotope. It has any more than 10 neutrons.
When we have eight, protons will go down seven protons. There's really nothing stable that has more than maybe eight neutrons. So the fact that we have 11 neutrons with ¹⁸₇N suggests that this is very unstable and
¹⁴₇N is the stable isotope of the pair.
<h3>
Stable and Unstable Nuclei</h3>
An atom is electrically neutral. It contains an equal number of positively charged protons and negatively charged electrons and their charges balance. The nucleus however contains only positively charged protons which are closely packed together in a very small volume (remember neutrons have no charge).
From the laws of physics (Coulomb’s Law) one would expect that the protons being of the same charge and so close together would exert strong repulsive forces on each other. The combined gravitational force from the protons and neutrons in a nucleus is insignificant as an attractive force because their masses are so tiny.
This implies there must be an additional attractive force similar in size to the electrostatic repulsion which holds the nucleus together.
Learn more about stable and unstable nuclei
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